Mixing
How many consecutive descending numbers in my number?
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2019 has come and probably everyone has noticed the peculiarity of this number: it's in fact composed by two sub-numbers (20 and 19) representing a sequence of consecutive descending numbers.
Challenge
Given a number x, return the length of the maximum sequence of consecutive, descending numbers that can be formed by taking sub-numbers of x.
Notes :
- sub-numbers cannot contain leading zeros (e.g.
1009cannot be split into10,09) - consecutive and descending means that a number in the sequence must be equal to the previous number -1, or $n_{i+1} = n_{i}-1$ (e.g.
52cannot be split into5,2because5and2are not consecutive,2 ≠ 5 - 1) - the sequence must be obtained by using the full number, e.g. in
7321you can't discard7and get the sequence3,2,1
- only one sequence can be obtained from the number, e.g.
3211098cannot be split into two sequences3,2,1and10,9,8
Input
- An integer number (
>= 0) : can be a number, or a string, or list of digits
Output
- A single integer given the maximum number of decreasing sub-numbers (note that the lower-bound of this number is
1, i.e. a number is composed by itself in a descending sequence of length one)
Examples :
2019 --> 20,19 --> output : 2
201200199198 --> 201,200,199,198 --> output : 4
3246 --> 3246 --> output : 1
87654 --> 8,7,6,5,4 --> output : 5
123456 --> 123456 --> output : 1
1009998 --> 100,99,98 --> output : 3
100908 --> 100908 --> output : 1
1110987 --> 11,10,9,8,7 --> output : 5
210 --> 2,1,0 --> output : 3
1 --> 1 --> output : 1
0 --> 0 --> output : 1
312 --> 312 --> output : 1
191 --> 191 --> output : 1
General rules:
- This is code-golf, so shortest answer in bytes wins.
Don't let code-golf languages discourage you from posting answers with non-codegolfing languages. Try to come up with an as short as possible answer for 'any' programming language.
Standard rules apply for your answer with default I/O rules, so you are allowed to use STDIN/STDOUT, functions/method with the proper parameters and return-type, full programs. Your call.
Default Loopholes are forbidden.- If possible, please add a link with a test for your code (i.e. TIO).
- Also, adding an explanation for your answer is highly recommended.
code-golf
edited Jan 5 at 1:08
Veskah
87614
asked Jan 4 at 10:48
digEmAlldigEmAll
3,009414
$endgroup$
add a comment |
$begingroup$
2019 has come and probably everyone has noticed the peculiarity of this number: it's in fact composed by two sub-numbers (20 and 19) representing a sequence of consecutive descending numbers.
Challenge
Given a number x, return the length of the maximum sequence of consecutive, descending numbers that can be formed by taking sub-numbers of x.
Notes :
- sub-numbers cannot contain leading zeros (e.g.
1009cannot be split into10,09) - consecutive and descending means that a number in the sequence must be equal to the previous number -1, or $n_{i+1} = n_{i}-1$ (e.g.
52cannot be split into5,2because5and2are not consecutive,2 ≠ 5 - 1) - the sequence must be obtained by using the full number, e.g. in
7321you can't discard7and get the sequence3,2,1
- only one sequence can be obtained from the number, e.g.
3211098cannot be split into two sequences3,2,1and10,9,8
Input
- An integer number (
>= 0) : can be a number, or a string, or list of digits
Output
- A single integer given the maximum number of decreasing sub-numbers (note that the lower-bound of this number is
1, i.e. a number is composed by itself in a descending sequence of length one)
Examples :
2019 --> 20,19 --> output : 2
201200199198 --> 201,200,199,198 --> output : 4
3246 --> 3246 --> output : 1
87654 --> 8,7,6,5,4 --> output : 5
123456 --> 123456 --> output : 1
1009998 --> 100,99,98 --> output : 3
100908 --> 100908 --> output : 1
1110987 --> 11,10,9,8,7 --> output : 5
210 --> 2,1,0 --> output : 3
1 --> 1 --> output : 1
0 --> 0 --> output : 1
312 --> 312 --> output : 1
191 --> 191 --> output : 1
General rules:
- This is code-golf, so shortest answer in bytes wins.
Don't let code-golf languages discourage you from posting answers with non-codegolfing languages. Try to come up with an as short as possible answer for 'any' programming language.
Standard rules apply for your answer with default I/O rules, so you are allowed to use STDIN/STDOUT, functions/method with the proper parameters and return-type, full programs. Your call.
Default Loopholes are forbidden.- If possible, please add a link with a test for your code (i.e. TIO).
- Also, adding an explanation for your answer is highly recommended.
code-golf
edited Jan 5 at 1:08
Veskah
87614
asked Jan 4 at 10:48
digEmAlldigEmAll
3,009414
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1
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Migrated from sandbox : codegolf.meta.stackexchange.com/questions/2140/…
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– digEmAll
Jan 4 at 10:49
1
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Is the test case210 -> 2,1,0wrong (same with0 -> 0)? The tasks says "sub-numbers cannot contain leading zeros", is zero a special case?
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– BMO
Jan 4 at 17:30
2
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@BMO: well, here the topic is kinda phylosofical... :D to me, 0 is a number with no (useless) leading zero, so yes zero is a special case
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– digEmAll
Jan 4 at 17:39
2
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would you call these... condescending numbers? xD sorry that was like not even funny
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– HyperNeutrino
Jan 6 at 3:03
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Sorry, deleted my comment in which I asked about212019. Seems I didn't read all the rules.
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– cyclaminist
Jan 12 at 21:49
add a comment |
$begingroup$
2019 has come and probably everyone has noticed the peculiarity of this number: it's in fact composed by two sub-numbers (20 and 19) representing a sequence of consecutive descending numbers.
Challenge
Given a number x, return the length of the maximum sequence of consecutive, descending numbers that can be formed by taking sub-numbers of x.
Notes :
- sub-numbers cannot contain leading zeros (e.g.
1009cannot be split into10,09) - consecutive and descending means that a number in the sequence must be equal to the previous number -1, or $n_{i+1} = n_{i}-1$ (e.g.
52cannot be split into5,2because5and2are not consecutive,2 ≠ 5 - 1) - the sequence must be obtained by using the full number, e.g. in
7321you can't discard7and get the sequence3,2,1
- only one sequence can be obtained from the number, e.g.
3211098cannot be split into two sequences3,2,1and10,9,8
Input
- An integer number (
>= 0) : can be a number, or a string, or list of digits
Output
- A single integer given the maximum number of decreasing sub-numbers (note that the lower-bound of this number is
1, i.e. a number is composed by itself in a descending sequence of length one)
Examples :
2019 --> 20,19 --> output : 2
201200199198 --> 201,200,199,198 --> output : 4
3246 --> 3246 --> output : 1
87654 --> 8,7,6,5,4 --> output : 5
123456 --> 123456 --> output : 1
1009998 --> 100,99,98 --> output : 3
100908 --> 100908 --> output : 1
1110987 --> 11,10,9,8,7 --> output : 5
210 --> 2,1,0 --> output : 3
1 --> 1 --> output : 1
0 --> 0 --> output : 1
312 --> 312 --> output : 1
191 --> 191 --> output : 1
General rules:
- This is code-golf, so shortest answer in bytes wins.
Don't let code-golf languages discourage you from posting answers with non-codegolfing languages. Try to come up with an as short as possible answer for 'any' programming language.
Standard rules apply for your answer with default I/O rules, so you are allowed to use STDIN/STDOUT, functions/method with the proper parameters and return-type, full programs. Your call.
Default Loopholes are forbidden.- If possible, please add a link with a test for your code (i.e. TIO).
- Also, adding an explanation for your answer is highly recommended.
code-golf
edited Jan 5 at 1:08
Veskah
87614
asked Jan 4 at 10:48
digEmAlldigEmAll
3,009414
$endgroup$
2019 has come and probably everyone has noticed the peculiarity of this number: it's in fact composed by two sub-numbers (20 and 19) representing a sequence of consecutive descending numbers.
Challenge
Given a number x, return the length of the maximum sequence of consecutive, descending numbers that can be formed by taking sub-numbers of x.
Notes :
- sub-numbers cannot contain leading zeros (e.g.
1009cannot be split into10,09) - consecutive and descending means that a number in the sequence must be equal to the previous number -1, or $n_{i+1} = n_{i}-1$ (e.g.
52cannot be split into5,2because5and2are not consecutive,2 ≠ 5 - 1) - the sequence must be obtained by using the full number, e.g. in
7321you can't discard7and get the sequence3,2,1
- only one sequence can be obtained from the number, e.g.
3211098cannot be split into two sequences3,2,1and10,9,8
Input
- An integer number (
>= 0) : can be a number, or a string, or list of digits
Output
- A single integer given the maximum number of decreasing sub-numbers (note that the lower-bound of this number is
1, i.e. a number is composed by itself in a descending sequence of length one)
Examples :
2019 --> 20,19 --> output : 2
201200199198 --> 201,200,199,198 --> output : 4
3246 --> 3246 --> output : 1
87654 --> 8,7,6,5,4 --> output : 5
123456 --> 123456 --> output : 1
1009998 --> 100,99,98 --> output : 3
100908 --> 100908 --> output : 1
1110987 --> 11,10,9,8,7 --> output : 5
210 --> 2,1,0 --> output : 3
1 --> 1 --> output : 1
0 --> 0 --> output : 1
312 --> 312 --> output : 1
191 --> 191 --> output : 1
General rules:
- This is code-golf, so shortest answer in bytes wins.
Don't let code-golf languages discourage you from posting answers with non-codegolfing languages. Try to come up with an as short as possible answer for 'any' programming language.
Standard rules apply for your answer with default I/O rules, so you are allowed to use STDIN/STDOUT, functions/method with the proper parameters and return-type, full programs. Your call.
Default Loopholes are forbidden.- If possible, please add a link with a test for your code (i.e. TIO).
- Also, adding an explanation for your answer is highly recommended.
code-golf
code-golf
edited Jan 5 at 1:08
Veskah
87614
asked Jan 4 at 10:48
digEmAlldigEmAll
3,009414
edited Jan 5 at 1:08
Veskah
87614
asked Jan 4 at 10:48
digEmAlldigEmAll
3,009414
edited Jan 5 at 1:08
Veskah
87614
edited Jan 5 at 1:08
Veskah
87614
edited Jan 5 at 1:08
Veskah
87614
87614
asked Jan 4 at 10:48
digEmAlldigEmAll
3,009414
asked Jan 4 at 10:48
digEmAlldigEmAll
3,009414
asked Jan 4 at 10:48
digEmAlldigEmAll
3,009414
3,009414
1
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Migrated from sandbox : codegolf.meta.stackexchange.com/questions/2140/…
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– digEmAll
Jan 4 at 10:49
1
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Is the test case210 -> 2,1,0wrong (same with0 -> 0)? The tasks says "sub-numbers cannot contain leading zeros", is zero a special case?
$endgroup$
– BMO
Jan 4 at 17:30
2
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@BMO: well, here the topic is kinda phylosofical... :D to me, 0 is a number with no (useless) leading zero, so yes zero is a special case
$endgroup$
– digEmAll
Jan 4 at 17:39
2
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would you call these... condescending numbers? xD sorry that was like not even funny
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– HyperNeutrino
Jan 6 at 3:03
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Sorry, deleted my comment in which I asked about212019. Seems I didn't read all the rules.
$endgroup$
– cyclaminist
Jan 12 at 21:49
add a comment |
1
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Migrated from sandbox : codegolf.meta.stackexchange.com/questions/2140/…
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– digEmAll
Jan 4 at 10:49
1
$begingroup$
Is the test case210 -> 2,1,0wrong (same with0 -> 0)? The tasks says "sub-numbers cannot contain leading zeros", is zero a special case?
$endgroup$
– BMO
Jan 4 at 17:30
2
$begingroup$
@BMO: well, here the topic is kinda phylosofical... :D to me, 0 is a number with no (useless) leading zero, so yes zero is a special case
$endgroup$
– digEmAll
Jan 4 at 17:39
2
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would you call these... condescending numbers? xD sorry that was like not even funny
$endgroup$
– HyperNeutrino
Jan 6 at 3:03
$begingroup$
Sorry, deleted my comment in which I asked about212019. Seems I didn't read all the rules.
$endgroup$
– cyclaminist
Jan 12 at 21:49
1
1
$begingroup$
Migrated from sandbox : codegolf.meta.stackexchange.com/questions/2140/…
$endgroup$
– digEmAll
Jan 4 at 10:49
$begingroup$
Migrated from sandbox : codegolf.meta.stackexchange.com/questions/2140/…
$endgroup$
– digEmAll
Jan 4 at 10:49
1
1
$begingroup$
Is the test case
210 -> 2,1,0 wrong (same with 0 -> 0)? The tasks says "sub-numbers cannot contain leading zeros", is zero a special case?$endgroup$
– BMO
Jan 4 at 17:30
$begingroup$
Is the test case
210 -> 2,1,0 wrong (same with 0 -> 0)? The tasks says "sub-numbers cannot contain leading zeros", is zero a special case?$endgroup$
– BMO
Jan 4 at 17:30
2
2
$begingroup$
@BMO: well, here the topic is kinda phylosofical... :D to me, 0 is a number with no (useless) leading zero, so yes zero is a special case
$endgroup$
– digEmAll
Jan 4 at 17:39
$begingroup$
@BMO: well, here the topic is kinda phylosofical... :D to me, 0 is a number with no (useless) leading zero, so yes zero is a special case
$endgroup$
– digEmAll
Jan 4 at 17:39
2
2
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would you call these... condescending numbers? xD sorry that was like not even funny
$endgroup$
– HyperNeutrino
Jan 6 at 3:03
$begingroup$
would you call these... condescending numbers? xD sorry that was like not even funny
$endgroup$
– HyperNeutrino
Jan 6 at 3:03
$begingroup$
Sorry, deleted my comment in which I asked about
212019. Seems I didn't read all the rules.$endgroup$
– cyclaminist
Jan 12 at 21:49
$begingroup$
Sorry, deleted my comment in which I asked about
212019. Seems I didn't read all the rules.$endgroup$
– cyclaminist
Jan 12 at 21:49
add a comment |
16 Answers
16
active
oldest
votes
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JavaScript (ES6), 56 bytes
A port of ArBo's Python answer is significantly shorter. However, it fails on some test cases because of too much recursion.
f=(n,a=0,c=0,s)=>a<0?f(n,a-~c):n==s?c:f(n,--a,c+1,[s]+a)
Try it online!
JavaScript (ES6), 66 bytes
Takes input as a string.
f=(s,n=x='',o=p=n,i=0)=>s[i++]?o==s?i:f(s,--n,o+n,i):f(s,p+s[x++])
Try it online!
Commented
f = ( // f = recursive function taking:
s, // s = input number, as a string
n = // n = counter
x = '', // x = position of the next digit to be added to p
o = p = n, // o = generated output; p = prefix
i = 0 // i = number of consecutive descending numbers
) => //
s[i++] ? // increment i; if s[i] was defined:
o == s ? // if o is matching s:
i // stop recursion and return i
: // else:
f( // do a recursive call with:
s, // s unchanged
--n, // n - 1
o + n, // (n - 1) appended to o
i // i unchanged (but it was incremented above)
) // end of recursive call
: // else:
f( // this is a dead end; try again with one more digit in the prefix:
s, // s unchanged
p + s[x++] // increment x and append the next digit to p
) // end of recursive call
answered Jan 4 at 11:25
ArnauldArnauld
73.5k689309
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54 bytes by implementing the changes to my code
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– ArBo
Jan 15 at 15:50
add a comment |
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Jelly, 15 9 bytes
Bugfix thanks to Dennis
ŻṚẆDfŒṖẈṀ
Try it online! (even 321 takes half a minute since the code is at least $O(N^2)$)
How?
ŻṚẆDfŒṖẈṀ - Link: integer, n
Ż - [0..n]
Ṛ - reverse
Ẇ - all contiguous slices (of implicit range(n)) = [[n],...,[2],[1],[0],[n,n-1],...,[2,1],[1,0],...,[n,n-1,n-2,...,2,1,0]]
D - to decimal (vectorises)
ŒṖ - partitions of (implicit decimal digits of) n
f - filter discard from left if in right
Ẉ - length of each
Ṁ - maximum
answered Jan 4 at 12:19
Jonathan AllanJonathan Allan
51.2k535166
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add a comment |
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Perl 6, 43 41 40 bytes
-1 byte thanks to nwellnhof
{/(<-[0]>.*?|0)+<?{[==] 1..*Z+$0}>/;+$0}
Try it online!
Regex based solution. I'm trying to come up with a better way to match from a descending list instead, but Perl 6 doesn't do partitions well
Explanation:
{ } # Anonymous code block
/ /; # Match in the input
<-[0]>.*? # Non-greedy number not starting with 0
|0 # Or 0
( )+ # Repeatedly for the rest of the number
<?{ }> # Where
1..*Z+$0 # Each matched number plus the ascending numbers
# For example 1,2,3 Z+ 9,8,7 is 10,10,10
[==] # Are all equal
+$0 # Return the length of the list
answered Jan 4 at 12:02
Jo KingJo King
21.4k248110
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40 bytes
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– nwellnhof
Jan 6 at 13:37
add a comment |
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Python 3, 232 228 187 181 180 150 149 bytes
-1 thanks to @Jonathan Frech
e=enumerate
t=int
h=lambda n,s=1:max([1]+[i-len(n[j:])and h(n[j:],s+1)or s+1for j,_ in e(n)for i,_ in e(n[:j],1)if(t(n[:j])-t(n[j:j+i])==1)*t(n[0])])
Try it online!
Initial ungolfed code:
def count_consecutives(left, right, so_far=1):
for i,_ in enumerate(left, start=1):
left_part_of_right, right_part_of_right = right[:i], right[i:]
if (int(left) - int(left_part_of_right)) == 1:
if i == len(right):
return so_far + 1
return count_consecutives(left_part_of_right, right_part_of_right, so_far + 1)
return so_far
def how_many_consecutives(n):
for i, _ in enumerate(n):
left, right = n[:i], n[i:]
for j, _ in enumerate(left, start=1):
left_part_of_right = right[:j]
if int(left) - int(left_part_of_right) == 1 and int(n[i]) > 0:
return count_consecutives(left, right)
return 1
answered Jan 5 at 16:39
NishiokaNishioka
1213
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1
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s+1 forcan bes+1for,(t(n[:j])-t(n[j:j+i])==1)*t(n[0])can possibly bet(n[:j])-t(n[j:j+i])==1>=t(n[0]).
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– Jonathan Frech
Jan 6 at 14:15
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It seems that second suggestion doesn't work though it wouldn't bring anything because then you need space to separate expression fromif.
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– Nishioka
Jan 6 at 14:34
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True ... alternative 149.
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– Jonathan Frech
Jan 6 at 15:03
add a comment |
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Python 2, 78 74 73 bytes
l=lambda n,a=0,c=0,s="":c*(n==s)or a and l(n,a-1,c+1,s+`a-1`)or l(n,a-~c)
Try it online!
-1 byte thanks to Arnauld
Takes input as a string. The program rather quickly runs into Python's recursion depth limit, but it can finish most of the test cases.
How it works
l=lambda n, # The input number, in the form of a string
a=0, # The program will attempt to reconstruct n by
# building a string by pasting decreasing
# numbers, stored in a, after each other.
c=0, # A counter of the amount of numbers
s="": # The current constructed string
c*(n==s) # Return the counter if s matches n
or # Else
a and l(n,a-1,c+1,s+`a-1`) # If a is not yet zero, paste a-1 after s
or # Else
l(n,a-~c) # Start again, from one higher than last time
answered Jan 14 at 20:23
ArBoArBo
32115
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1
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Nice answer!a+c+1can be shortened toa-~c.
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– Arnauld
Jan 14 at 21:01
add a comment |
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05AB1E, 10 bytes
ÝRŒʒJQ}€gà
Extremely slow, so the TIO below only works for test cases below 750..
Try it online.
Explanation:
Ý # Create a list in the range [0, (implicit) input]
# i.e. 109 → [0,1,2,...,107,108,109]
R # Reverse it
# i.e. [0,1,2,...,107,108,109] → [109,108,107,...,2,1,0]
Π# Get all possible sublists of this list
# i.e. [109,108,107,...,2,1,0]
# → [[109],[109,108],[109,108,107],...,[2,1,0],[1],[1,0],[0]]
ʒ } # Filter it by:
J # Where the sublist joined together
# i.e. [10,9] → "109"
# i.e. [109,108,107] → "109108107"
Q # Are equal to the (implicit) input
# i.e. 109 and "109" → 1 (truthy)
# i.e. 109 and "109108107" → 0 (falsey)
€g # After filtering, take the length of each remaining inner list
# i.e. [[109],[[10,9]] → [1,2]
à # And only leave the maximum length (which is output implicitly)
# i.e. [1,2] → 2
answered Jan 4 at 12:18
Kevin CruijssenKevin Cruijssen
36.5k555192
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2
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Code golf - where adding 1 byte to your program to go fromn!ton lg njust isn't worth it.
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– corsiKa
Jan 5 at 20:49
add a comment |
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Pyth, 16 bytes
lef!.EhM.+vMT./z
Try it online here, or verify all the test cases at once here.
lef!.EhM.+vMT./z Implicit: z=input as string
./z Get all divisions of z into disjoint substrings
f Filter the above, as T, keeping those where the following is truthy:
vMT Parse each substring as an int
.+ Get difference between each pair
hM Increment each
!.E Are all elements 0? { NOT(ANY(...)) }
e Take the last element of the filtered divisions
Divisions are generated with fewest substrings first, so last remaining division is also the longest
l Length of the above, implicit print
answered Jan 4 at 12:52
SokSok
3,637723
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add a comment |
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Charcoal, 26 bytes
Ｆ⊕ＬθＦ⊕Ｌθ⊞υ⭆κ⁻Ｉ…θιλＩ﹪⌕υθ⊕Ｌθ
Try it online! Link is to verbose version of code. Explanation:
Ｆ⊕Ｌθ
Loop i from 0 to the length of the input.
Ｆ⊕Ｌθ
Loop k from 0 to the length of the input.
⊞υ⭆κ⁻Ｉ…θ⊕ιλ
Calculate the first k numbers in the descending sequence starting from the number given by the first i digits of the input, concatenate them, and accumulate each resulting string in the predefined empty list.
Ｉ﹪⌕υθ⊕Ｌθ
Find the position of the first matching copy of the input and reduce it modulo 1 more than the length of the input.
Example: For an input of 2019 the following strings are generated:
0
1 0
2 0-1
3 0-1-2
4 0-1-2-3
5
6 2
7 21
8 210
9 210-1
10
11 20
12 2019
13 201918
14 20191817
15
16 201
17 201200
18 201200199
19 201200199198
20
21 2019
22 20192018
23 201920182017
24 2019201820172016
2019 is then found at index 12, which is reduced modulo 5 to give 2, the desired answer.
answered Jan 4 at 21:20
NeilNeil
79.9k744178
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add a comment |
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Haskell, 87 bytes
maximum.map length.(0#)
a#(b:c)=[a:x|c==||b>0,x<-b#c,a==x!!0+1]++(10*a+b)#c
a#b=[[a]]
Input is a list of digits.
Try it online!
Function # builds a list of all possible splits by looking at both
- prepending the current number
ato all splits returned by a recursive call with the rest of the input (x<-b#c), but only if the next number not zero (b>0) (or it's the last number in the input (c==)) andais one greater than the first number of the respective previous splitx(a==x!!0+1).
and
- appending the next digit
bfrom the input list to the current numberaand going on with the rest of the input ((10*a+b)#c)
Base case is when the input list is empty (i.e. does not pattern match (b:c)). The recursion starts with the current number a being 0 ((0#)), which never hits the first branch (prepending a to all previous splits), because it will never be greater than any number of the splits.
Take the length of each split and find the maximum (maximum.map length).
A variant with also 87 bytes:
fst.maximum.(0#)
a#(b:c)=[(r+1,a)|c==||b>0,(r,x)<-b#c,a==x+1]++(10*a+b)#c
a#b=[(1,a)]
which basically works the same way, but instead of keeping the whole split in a list, it only keeps a pair (r,x) of the length of the split r an the first number in the split x.
answered Jan 5 at 15:00
niminimi
31.5k32185
$endgroup$
add a comment |
$begingroup$
Python 3, 302 282 271 bytes
-10 bytes thanks to the tip by @ElPedro.
Takes input as a string. Basically, it takes increasing larger slices of the number from the left, and sees if for that slice of the number a sequence can be formed using all the numbers.
R=range
I=int
L=len
def g(n,m,t=1):
for i in R(1,L(m)+1):
if I(m)==I(n[:i])+1:
if i==L(n):return-~t
return g(n[i:],n[:i],t+1)
return 1
def f(n):
for i in R(L(n)):
x=n[:i]
for j in R(1,L(x)+1):
if (I(x)==I(n[i:i+j])+1)*I(n[i]):return g(n[i:],x)
return 1
Try it online!
answered Jan 4 at 20:39
Neil A.Neil A.
1,218120
$endgroup$
1
$begingroup$
Since you are usingrange3 times you can defineR=rangeoutside of both functions and then useR(whatever)instead ofrange(whatever)to save 4 bytes.
$endgroup$
– ElPedro
Jan 5 at 17:48
add a comment |
$begingroup$
Japt, 27 bytes
ò pÊÔpÊqÊfl²i1Uì q"l?"¹ÌèÊÉ
Try it online! or Check most test cases
This doesn't score well, but it uses a unique method and there might be room to golf it a lot more. It also performs well enough that all test cases other than 201200199198 avoid timing out.
Explanation:
ò #Get the range [0...input]
pÊ #Add an "l" to the end
Ô #Reverse it
pÊ #Add an "l" to the end
qÊ #Add an "l" between each number and turn to a string
f ¹ #Find the substrings that match this regex:
l² # The string "ll"
i1 # With this inserted between the "l"s:
Uì # All the digits of the input
q"l?" # With optional spaces between each one
Ì #Get the last match
èÊ #Count the number of "l"s
É #Subtract 1
answered Jan 4 at 20:02
Kamil DrakariKamil Drakari
3,131416
$endgroup$
$begingroup$
I think this works for 27.
$endgroup$
– Shaggy
Jan 5 at 10:30
$begingroup$
25 bytes
$endgroup$
– Shaggy
Jan 5 at 10:43
$begingroup$
@Shaggy both of those fail on input21201because they don't enforce that the sequence end aligns correctly (from my original version the "ends with a comma" line). This or this alternative works.
$endgroup$
– Kamil Drakari
Jan 5 at 15:38
$begingroup$
Ah, OK. In that case: 26 bytes
$endgroup$
– Shaggy
Jan 5 at 15:46
$begingroup$
@Shaggy That and the 28 byte solutions I had fail on210because there isn't a delimiter after 0. Here's a fixed 28 byte that works.
$endgroup$
– Kamil Drakari
Jan 6 at 1:35
|
show 1 more comment
$begingroup$
Jelly, 11 bytes
ŒṖḌ’DɗƑƇẈṀ
Byte for byte, no match for the other Jelly solution, but this one should be roughly $Oleft(n^{0.3}right)$.
Try it online!
How it works
ŒṖḌ’DɗƑƇẈṀ Main link. Argument: n (integer)
ŒṖ Yield all partitions of n's digit list in base 10.
Ƈ Comb; keep only partitions for which the link to the left returns 1.
Ƒ Fixed; yield 1 if calling the link to the left returns its argument.
Cumulatively reduce the partition by the link to the left.
ɗ Combine the three links to the left into a dyadic chain.
Ḍ Undecimal; convert a digit list into an integer.
’ Decrement the result.
D Decimal; convert the integer back to a digit list.
answered Jan 4 at 14:02
Dennis♦Dennis
187k32297737
$endgroup$
add a comment |
$begingroup$
Haskell, 65 bytes
f i=[y|x<-[0..],y<-[1..length i],i==(show=<<[x+y-1,x+y-2..x])]!!0
Input is a string.
Try it online!
Completely different to my other answer. A simple brute force that tries all lists of consecutive descending numbers until it finds one that equals the input list.
If we limit the input number to 64-bit integers, we can save 6 bytes by looping y through [1..19], because the largest 64-bit integer has 19 digits and there's no need to test lists with more elements.
Haskell, 59 bytes
f i=[y|x<-[0..],y<-[1..19],i==(show=<<[x+y-1,x+y-2..x])]!!0
Try it online!
answered Jan 5 at 16:02
niminimi
31.5k32185
$endgroup$
add a comment |
$begingroup$
Python 2, 95 bytes
lambda n:max(j-i for j in range(n+1)for i in range(-1,j)if''.join(map(str,range(j,i,-1)))==`n`)
Another slow, brute-force solution.
Try it online!
answered Jan 5 at 20:32
Dennis♦Dennis
187k32297737
$endgroup$
add a comment |
$begingroup$
Dyalog APL, 138 bytes
Bit of a mouthful, but it works fast for big numbers too. If you try it online, prefix the dfn by ⎕← and provide input on the right as a list of digits.
{⌈/⍵((≢⊂)×1∧.=2-/10⊥¨⊂)⍨⍤1⊢1,{⍬≡1↓⍵:↑⍬1⋄0=⊃⍵:0,∇1↓⍵⋄↑,0 1∘.,⊂⍤1∇1↓⍵}1↓⍵}
Explanation
First, the inner dfn on the right which recursively constructs a list of possible ways to partition (with ⊂) the list of digits. For example 1 0 1 0 ⊂ 2 0 1 9 returns the nested vector (2 0)(1 9).
{
⍬≡1↓⍵: ↑⍬1 ⍝ Edge case: If ⍵ is singleton list, return the column matrix (0 1)
0=⊃⍵: 0,∇1↓⍵ ⍝ If head of ⍵ is 0, return 0 catenated to this dfn called on tail ⍵
↑,0 1∘.,⊂⍤1∇1↓⍵ ⍝ Finds 1 cat recursive call on tail ⍵ and 0 cat recursive call on ⍵.
} ⍝ Makes a matrix with a row for each possibility.
We use 1, to add a column of 1s at the start and end up with a matrix of valid partitions for ⍵.
Now the function train on the left in parens. Due to ⍨ the left argument to the train is the row of the partitions matrix and the right argument is the user input.
The train is a pile of forks with an atop as the far left tine.
((≢⊂)×1∧.=2-/10⊥¨⊂)⍨ ⍝ ⍨ swaps left and right arguments of the train.
⊂ ⍝ Partition ⍵ according to ⍺.
10⊥¨ ⍝ Decode each partition (turns strings of digits into numbers)
2-/ ⍝ Difference between adjacent cells
1∧.= ⍝ All equal 1?
⊂ ⍝ Partition ⍵ according to ⍺ again
≢ ⍝ Number of cells (ie number of partitions)
× ⍝ Multiply.
If the partition creates a sequence of descending numbers, the train returns the length of the sequence. Otherwise zero.
⍤1⊢ applies the function train between the user input and each row of the partitions matrix, returning a value for each row of the matrix. ⊢ is necessary to disambiguate between operand to ⍤ and argument to the derived function of ⍤.
⌈/ finds the maximum.
Could find a shorter algorithm but I wanted to try this way which is the most direct and declarative I could think of.
answered Jan 12 at 21:39
akhmornakhmorn
312
$endgroup$
$begingroup$
Welcome to PPCG! This is an impressive first post!
$endgroup$
– Riker
Jan 12 at 21:47
add a comment |
$begingroup$
TSQL, 169 bytes
Note: this can only be executed when the input can be converted to an integer.
Recursive sql used for looping.
Golfed:
DECLARE @ varchar(max) = '1211109876';
WITH C as(SELECT left(@,row_number()over(order by 1/0))+0t,@+null z,0i
FROM spt_values UNION ALL
SELECT t-1,concat(z,t),i+1FROM C WHERE i<9)SELECT
max(i)FROM C WHERE z=@
Ungolfed:
DECLARE @ varchar(max) = '1211109876';
WITH C as
(
SELECT
left(@,row_number()over(order by 1/0))+0t,
@+null z,
0i
FROM
spt_values
UNION ALL
SELECT
t-1,
concat(z,t),
i+1
FROM C
WHERE i<9
)
SELECT max(i)
FROM C
WHERE z=@
Try it out
answered Jan 11 at 21:24
t-clausen.dkt-clausen.dk
1,834314
$endgroup$
add a comment |
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16 Answers
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16 Answers
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$begingroup$
JavaScript (ES6), 56 bytes
A port of ArBo's Python answer is significantly shorter. However, it fails on some test cases because of too much recursion.
f=(n,a=0,c=0,s)=>a<0?f(n,a-~c):n==s?c:f(n,--a,c+1,[s]+a)
Try it online!
JavaScript (ES6), 66 bytes
Takes input as a string.
f=(s,n=x='',o=p=n,i=0)=>s[i++]?o==s?i:f(s,--n,o+n,i):f(s,p+s[x++])
Try it online!
Commented
f = ( // f = recursive function taking:
s, // s = input number, as a string
n = // n = counter
x = '', // x = position of the next digit to be added to p
o = p = n, // o = generated output; p = prefix
i = 0 // i = number of consecutive descending numbers
) => //
s[i++] ? // increment i; if s[i] was defined:
o == s ? // if o is matching s:
i // stop recursion and return i
: // else:
f( // do a recursive call with:
s, // s unchanged
--n, // n - 1
o + n, // (n - 1) appended to o
i // i unchanged (but it was incremented above)
) // end of recursive call
: // else:
f( // this is a dead end; try again with one more digit in the prefix:
s, // s unchanged
p + s[x++] // increment x and append the next digit to p
) // end of recursive call
answered Jan 4 at 11:25
ArnauldArnauld
73.5k689309
$endgroup$
$begingroup$
54 bytes by implementing the changes to my code
$endgroup$
– ArBo
Jan 15 at 15:50
add a comment |
$begingroup$
JavaScript (ES6), 56 bytes
A port of ArBo's Python answer is significantly shorter. However, it fails on some test cases because of too much recursion.
f=(n,a=0,c=0,s)=>a<0?f(n,a-~c):n==s?c:f(n,--a,c+1,[s]+a)
Try it online!
JavaScript (ES6), 66 bytes
Takes input as a string.
f=(s,n=x='',o=p=n,i=0)=>s[i++]?o==s?i:f(s,--n,o+n,i):f(s,p+s[x++])
Try it online!
Commented
f = ( // f = recursive function taking:
s, // s = input number, as a string
n = // n = counter
x = '', // x = position of the next digit to be added to p
o = p = n, // o = generated output; p = prefix
i = 0 // i = number of consecutive descending numbers
) => //
s[i++] ? // increment i; if s[i] was defined:
o == s ? // if o is matching s:
i // stop recursion and return i
: // else:
f( // do a recursive call with:
s, // s unchanged
--n, // n - 1
o + n, // (n - 1) appended to o
i // i unchanged (but it was incremented above)
) // end of recursive call
: // else:
f( // this is a dead end; try again with one more digit in the prefix:
s, // s unchanged
p + s[x++] // increment x and append the next digit to p
) // end of recursive call
answered Jan 4 at 11:25
ArnauldArnauld
73.5k689309
$endgroup$
$begingroup$
54 bytes by implementing the changes to my code
$endgroup$
– ArBo
Jan 15 at 15:50
add a comment |
$begingroup$
JavaScript (ES6), 56 bytes
A port of ArBo's Python answer is significantly shorter. However, it fails on some test cases because of too much recursion.
f=(n,a=0,c=0,s)=>a<0?f(n,a-~c):n==s?c:f(n,--a,c+1,[s]+a)
Try it online!
JavaScript (ES6), 66 bytes
Takes input as a string.
f=(s,n=x='',o=p=n,i=0)=>s[i++]?o==s?i:f(s,--n,o+n,i):f(s,p+s[x++])
Try it online!
Commented
f = ( // f = recursive function taking:
s, // s = input number, as a string
n = // n = counter
x = '', // x = position of the next digit to be added to p
o = p = n, // o = generated output; p = prefix
i = 0 // i = number of consecutive descending numbers
) => //
s[i++] ? // increment i; if s[i] was defined:
o == s ? // if o is matching s:
i // stop recursion and return i
: // else:
f( // do a recursive call with:
s, // s unchanged
--n, // n - 1
o + n, // (n - 1) appended to o
i // i unchanged (but it was incremented above)
) // end of recursive call
: // else:
f( // this is a dead end; try again with one more digit in the prefix:
s, // s unchanged
p + s[x++] // increment x and append the next digit to p
) // end of recursive call
answered Jan 4 at 11:25
ArnauldArnauld
73.5k689309
$endgroup$
JavaScript (ES6), 56 bytes
A port of ArBo's Python answer is significantly shorter. However, it fails on some test cases because of too much recursion.
f=(n,a=0,c=0,s)=>a<0?f(n,a-~c):n==s?c:f(n,--a,c+1,[s]+a)
Try it online!
JavaScript (ES6), 66 bytes
Takes input as a string.
f=(s,n=x='',o=p=n,i=0)=>s[i++]?o==s?i:f(s,--n,o+n,i):f(s,p+s[x++])
Try it online!
Commented
f = ( // f = recursive function taking:
s, // s = input number, as a string
n = // n = counter
x = '', // x = position of the next digit to be added to p
o = p = n, // o = generated output; p = prefix
i = 0 // i = number of consecutive descending numbers
) => //
s[i++] ? // increment i; if s[i] was defined:
o == s ? // if o is matching s:
i // stop recursion and return i
: // else:
f( // do a recursive call with:
s, // s unchanged
--n, // n - 1
o + n, // (n - 1) appended to o
i // i unchanged (but it was incremented above)
) // end of recursive call
: // else:
f( // this is a dead end; try again with one more digit in the prefix:
s, // s unchanged
p + s[x++] // increment x and append the next digit to p
) // end of recursive call
answered Jan 4 at 11:25
ArnauldArnauld
73.5k689309
edited Jan 14 at 21:21
answered Jan 4 at 11:25
ArnauldArnauld
73.5k689309
answered Jan 4 at 11:25
ArnauldArnauld
73.5k689309
answered Jan 4 at 11:25
ArnauldArnauld
73.5k689309
73.5k689309
$begingroup$
54 bytes by implementing the changes to my code
$endgroup$
– ArBo
Jan 15 at 15:50
add a comment |
$begingroup$
54 bytes by implementing the changes to my code
$endgroup$
– ArBo
Jan 15 at 15:50
$begingroup$
54 bytes by implementing the changes to my code
$endgroup$
– ArBo
Jan 15 at 15:50
$begingroup$
54 bytes by implementing the changes to my code
$endgroup$
– ArBo
Jan 15 at 15:50
add a comment |
$begingroup$
Jelly, 15 9 bytes
Bugfix thanks to Dennis
ŻṚẆDfŒṖẈṀ
Try it online! (even 321 takes half a minute since the code is at least $O(N^2)$)
How?
ŻṚẆDfŒṖẈṀ - Link: integer, n
Ż - [0..n]
Ṛ - reverse
Ẇ - all contiguous slices (of implicit range(n)) = [[n],...,[2],[1],[0],[n,n-1],...,[2,1],[1,0],...,[n,n-1,n-2,...,2,1,0]]
D - to decimal (vectorises)
ŒṖ - partitions of (implicit decimal digits of) n
f - filter discard from left if in right
Ẉ - length of each
Ṁ - maximum
answered Jan 4 at 12:19
Jonathan AllanJonathan Allan
51.2k535166
$endgroup$
add a comment |
$begingroup$
Jelly, 15 9 bytes
Bugfix thanks to Dennis
ŻṚẆDfŒṖẈṀ
Try it online! (even 321 takes half a minute since the code is at least $O(N^2)$)
How?
ŻṚẆDfŒṖẈṀ - Link: integer, n
Ż - [0..n]
Ṛ - reverse
Ẇ - all contiguous slices (of implicit range(n)) = [[n],...,[2],[1],[0],[n,n-1],...,[2,1],[1,0],...,[n,n-1,n-2,...,2,1,0]]
D - to decimal (vectorises)
ŒṖ - partitions of (implicit decimal digits of) n
f - filter discard from left if in right
Ẉ - length of each
Ṁ - maximum
answered Jan 4 at 12:19
Jonathan AllanJonathan Allan
51.2k535166
$endgroup$
add a comment |
$begingroup$
Jelly, 15 9 bytes
Bugfix thanks to Dennis
ŻṚẆDfŒṖẈṀ
Try it online! (even 321 takes half a minute since the code is at least $O(N^2)$)
How?
ŻṚẆDfŒṖẈṀ - Link: integer, n
Ż - [0..n]
Ṛ - reverse
Ẇ - all contiguous slices (of implicit range(n)) = [[n],...,[2],[1],[0],[n,n-1],...,[2,1],[1,0],...,[n,n-1,n-2,...,2,1,0]]
D - to decimal (vectorises)
ŒṖ - partitions of (implicit decimal digits of) n
f - filter discard from left if in right
Ẉ - length of each
Ṁ - maximum
answered Jan 4 at 12:19
Jonathan AllanJonathan Allan
51.2k535166
$endgroup$
Jelly, 15 9 bytes
Bugfix thanks to Dennis
ŻṚẆDfŒṖẈṀ
Try it online! (even 321 takes half a minute since the code is at least $O(N^2)$)
How?
ŻṚẆDfŒṖẈṀ - Link: integer, n
Ż - [0..n]
Ṛ - reverse
Ẇ - all contiguous slices (of implicit range(n)) = [[n],...,[2],[1],[0],[n,n-1],...,[2,1],[1,0],...,[n,n-1,n-2,...,2,1,0]]
D - to decimal (vectorises)
ŒṖ - partitions of (implicit decimal digits of) n
f - filter discard from left if in right
Ẉ - length of each
Ṁ - maximum
answered Jan 4 at 12:19
Jonathan AllanJonathan Allan
51.2k535166
edited Jan 4 at 20:42
answered Jan 4 at 12:19
Jonathan AllanJonathan Allan
51.2k535166
answered Jan 4 at 12:19
Jonathan AllanJonathan Allan
51.2k535166
answered Jan 4 at 12:19
Jonathan AllanJonathan Allan
51.2k535166
51.2k535166
add a comment |
add a comment |
$begingroup$
Perl 6, 43 41 40 bytes
-1 byte thanks to nwellnhof
{/(<-[0]>.*?|0)+<?{[==] 1..*Z+$0}>/;+$0}
Try it online!
Regex based solution. I'm trying to come up with a better way to match from a descending list instead, but Perl 6 doesn't do partitions well
Explanation:
{ } # Anonymous code block
/ /; # Match in the input
<-[0]>.*? # Non-greedy number not starting with 0
|0 # Or 0
( )+ # Repeatedly for the rest of the number
<?{ }> # Where
1..*Z+$0 # Each matched number plus the ascending numbers
# For example 1,2,3 Z+ 9,8,7 is 10,10,10
[==] # Are all equal
+$0 # Return the length of the list
answered Jan 4 at 12:02
Jo KingJo King
21.4k248110
$endgroup$
$begingroup$
40 bytes
$endgroup$
– nwellnhof
Jan 6 at 13:37
add a comment |
$begingroup$
Perl 6, 43 41 40 bytes
-1 byte thanks to nwellnhof
{/(<-[0]>.*?|0)+<?{[==] 1..*Z+$0}>/;+$0}
Try it online!
Regex based solution. I'm trying to come up with a better way to match from a descending list instead, but Perl 6 doesn't do partitions well
Explanation:
{ } # Anonymous code block
/ /; # Match in the input
<-[0]>.*? # Non-greedy number not starting with 0
|0 # Or 0
( )+ # Repeatedly for the rest of the number
<?{ }> # Where
1..*Z+$0 # Each matched number plus the ascending numbers
# For example 1,2,3 Z+ 9,8,7 is 10,10,10
[==] # Are all equal
+$0 # Return the length of the list
answered Jan 4 at 12:02
Jo KingJo King
21.4k248110
$endgroup$
$begingroup$
40 bytes
$endgroup$
– nwellnhof
Jan 6 at 13:37
add a comment |
$begingroup$
Perl 6, 43 41 40 bytes
-1 byte thanks to nwellnhof
{/(<-[0]>.*?|0)+<?{[==] 1..*Z+$0}>/;+$0}
Try it online!
Regex based solution. I'm trying to come up with a better way to match from a descending list instead, but Perl 6 doesn't do partitions well
Explanation:
{ } # Anonymous code block
/ /; # Match in the input
<-[0]>.*? # Non-greedy number not starting with 0
|0 # Or 0
( )+ # Repeatedly for the rest of the number
<?{ }> # Where
1..*Z+$0 # Each matched number plus the ascending numbers
# For example 1,2,3 Z+ 9,8,7 is 10,10,10
[==] # Are all equal
+$0 # Return the length of the list
answered Jan 4 at 12:02
Jo KingJo King
21.4k248110
$endgroup$
Perl 6, 43 41 40 bytes
-1 byte thanks to nwellnhof
{/(<-[0]>.*?|0)+<?{[==] 1..*Z+$0}>/;+$0}
Try it online!
Regex based solution. I'm trying to come up with a better way to match from a descending list instead, but Perl 6 doesn't do partitions well
Explanation:
{ } # Anonymous code block
/ /; # Match in the input
<-[0]>.*? # Non-greedy number not starting with 0
|0 # Or 0
( )+ # Repeatedly for the rest of the number
<?{ }> # Where
1..*Z+$0 # Each matched number plus the ascending numbers
# For example 1,2,3 Z+ 9,8,7 is 10,10,10
[==] # Are all equal
+$0 # Return the length of the list
answered Jan 4 at 12:02
Jo KingJo King
21.4k248110
edited Jan 6 at 22:12
answered Jan 4 at 12:02
Jo KingJo King
21.4k248110
answered Jan 4 at 12:02
Jo KingJo King
21.4k248110
answered Jan 4 at 12:02
Jo KingJo King
21.4k248110
21.4k248110
$begingroup$
40 bytes
$endgroup$
– nwellnhof
Jan 6 at 13:37
add a comment |
$begingroup$
40 bytes
$endgroup$
– nwellnhof
Jan 6 at 13:37
$begingroup$
40 bytes
$endgroup$
– nwellnhof
Jan 6 at 13:37
$begingroup$
40 bytes
$endgroup$
– nwellnhof
Jan 6 at 13:37
add a comment |
$begingroup$
Python 3, 232 228 187 181 180 150 149 bytes
-1 thanks to @Jonathan Frech
e=enumerate
t=int
h=lambda n,s=1:max([1]+[i-len(n[j:])and h(n[j:],s+1)or s+1for j,_ in e(n)for i,_ in e(n[:j],1)if(t(n[:j])-t(n[j:j+i])==1)*t(n[0])])
Try it online!
Initial ungolfed code:
def count_consecutives(left, right, so_far=1):
for i,_ in enumerate(left, start=1):
left_part_of_right, right_part_of_right = right[:i], right[i:]
if (int(left) - int(left_part_of_right)) == 1:
if i == len(right):
return so_far + 1
return count_consecutives(left_part_of_right, right_part_of_right, so_far + 1)
return so_far
def how_many_consecutives(n):
for i, _ in enumerate(n):
left, right = n[:i], n[i:]
for j, _ in enumerate(left, start=1):
left_part_of_right = right[:j]
if int(left) - int(left_part_of_right) == 1 and int(n[i]) > 0:
return count_consecutives(left, right)
return 1
answered Jan 5 at 16:39
NishiokaNishioka
1213
$endgroup$
1
$begingroup$
s+1 forcan bes+1for,(t(n[:j])-t(n[j:j+i])==1)*t(n[0])can possibly bet(n[:j])-t(n[j:j+i])==1>=t(n[0]).
$endgroup$
– Jonathan Frech
Jan 6 at 14:15
$begingroup$
It seems that second suggestion doesn't work though it wouldn't bring anything because then you need space to separate expression fromif.
$endgroup$
– Nishioka
Jan 6 at 14:34
$begingroup$
True ... alternative 149.
$endgroup$
– Jonathan Frech
Jan 6 at 15:03
add a comment |
$begingroup$
Python 3, 232 228 187 181 180 150 149 bytes
-1 thanks to @Jonathan Frech
e=enumerate
t=int
h=lambda n,s=1:max([1]+[i-len(n[j:])and h(n[j:],s+1)or s+1for j,_ in e(n)for i,_ in e(n[:j],1)if(t(n[:j])-t(n[j:j+i])==1)*t(n[0])])
Try it online!
Initial ungolfed code:
def count_consecutives(left, right, so_far=1):
for i,_ in enumerate(left, start=1):
left_part_of_right, right_part_of_right = right[:i], right[i:]
if (int(left) - int(left_part_of_right)) == 1:
if i == len(right):
return so_far + 1
return count_consecutives(left_part_of_right, right_part_of_right, so_far + 1)
return so_far
def how_many_consecutives(n):
for i, _ in enumerate(n):
left, right = n[:i], n[i:]
for j, _ in enumerate(left, start=1):
left_part_of_right = right[:j]
if int(left) - int(left_part_of_right) == 1 and int(n[i]) > 0:
return count_consecutives(left, right)
return 1
answered Jan 5 at 16:39
NishiokaNishioka
1213
$endgroup$
1
$begingroup$
s+1 forcan bes+1for,(t(n[:j])-t(n[j:j+i])==1)*t(n[0])can possibly bet(n[:j])-t(n[j:j+i])==1>=t(n[0]).
$endgroup$
– Jonathan Frech
Jan 6 at 14:15
$begingroup$
It seems that second suggestion doesn't work though it wouldn't bring anything because then you need space to separate expression fromif.
$endgroup$
– Nishioka
Jan 6 at 14:34
$begingroup$
True ... alternative 149.
$endgroup$
– Jonathan Frech
Jan 6 at 15:03
add a comment |
$begingroup$
Python 3, 232 228 187 181 180 150 149 bytes
-1 thanks to @Jonathan Frech
e=enumerate
t=int
h=lambda n,s=1:max([1]+[i-len(n[j:])and h(n[j:],s+1)or s+1for j,_ in e(n)for i,_ in e(n[:j],1)if(t(n[:j])-t(n[j:j+i])==1)*t(n[0])])
Try it online!
Initial ungolfed code:
def count_consecutives(left, right, so_far=1):
for i,_ in enumerate(left, start=1):
left_part_of_right, right_part_of_right = right[:i], right[i:]
if (int(left) - int(left_part_of_right)) == 1:
if i == len(right):
return so_far + 1
return count_consecutives(left_part_of_right, right_part_of_right, so_far + 1)
return so_far
def how_many_consecutives(n):
for i, _ in enumerate(n):
left, right = n[:i], n[i:]
for j, _ in enumerate(left, start=1):
left_part_of_right = right[:j]
if int(left) - int(left_part_of_right) == 1 and int(n[i]) > 0:
return count_consecutives(left, right)
return 1
answered Jan 5 at 16:39
NishiokaNishioka
1213
$endgroup$
Python 3, 232 228 187 181 180 150 149 bytes
-1 thanks to @Jonathan Frech
e=enumerate
t=int
h=lambda n,s=1:max([1]+[i-len(n[j:])and h(n[j:],s+1)or s+1for j,_ in e(n)for i,_ in e(n[:j],1)if(t(n[:j])-t(n[j:j+i])==1)*t(n[0])])
Try it online!
Initial ungolfed code:
def count_consecutives(left, right, so_far=1):
for i,_ in enumerate(left, start=1):
left_part_of_right, right_part_of_right = right[:i], right[i:]
if (int(left) - int(left_part_of_right)) == 1:
if i == len(right):
return so_far + 1
return count_consecutives(left_part_of_right, right_part_of_right, so_far + 1)
return so_far
def how_many_consecutives(n):
for i, _ in enumerate(n):
left, right = n[:i], n[i:]
for j, _ in enumerate(left, start=1):
left_part_of_right = right[:j]
if int(left) - int(left_part_of_right) == 1 and int(n[i]) > 0:
return count_consecutives(left, right)
return 1
answered Jan 5 at 16:39
NishiokaNishioka
1213
edited Jan 6 at 14:38
answered Jan 5 at 16:39
NishiokaNishioka
1213
answered Jan 5 at 16:39
NishiokaNishioka
1213
answered Jan 5 at 16:39
NishiokaNishioka
1213
1213
1
$begingroup$
s+1 forcan bes+1for,(t(n[:j])-t(n[j:j+i])==1)*t(n[0])can possibly bet(n[:j])-t(n[j:j+i])==1>=t(n[0]).
$endgroup$
– Jonathan Frech
Jan 6 at 14:15
$begingroup$
It seems that second suggestion doesn't work though it wouldn't bring anything because then you need space to separate expression fromif.
$endgroup$
– Nishioka
Jan 6 at 14:34
$begingroup$
True ... alternative 149.
$endgroup$
– Jonathan Frech
Jan 6 at 15:03
add a comment |
1
$begingroup$
s+1 forcan bes+1for,(t(n[:j])-t(n[j:j+i])==1)*t(n[0])can possibly bet(n[:j])-t(n[j:j+i])==1>=t(n[0]).
$endgroup$
– Jonathan Frech
Jan 6 at 14:15
$begingroup$
It seems that second suggestion doesn't work though it wouldn't bring anything because then you need space to separate expression fromif.
$endgroup$
– Nishioka
Jan 6 at 14:34
$begingroup$
True ... alternative 149.
$endgroup$
– Jonathan Frech
Jan 6 at 15:03
1
1
$begingroup$
s+1 for can be s+1for, (t(n[:j])-t(n[j:j+i])==1)*t(n[0]) can possibly be t(n[:j])-t(n[j:j+i])==1>=t(n[0]).$endgroup$
– Jonathan Frech
Jan 6 at 14:15
$begingroup$
s+1 for can be s+1for, (t(n[:j])-t(n[j:j+i])==1)*t(n[0]) can possibly be t(n[:j])-t(n[j:j+i])==1>=t(n[0]).$endgroup$
– Jonathan Frech
Jan 6 at 14:15
$begingroup$
It seems that second suggestion doesn't work though it wouldn't bring anything because then you need space to separate expression from
if.$endgroup$
– Nishioka
Jan 6 at 14:34
$begingroup$
It seems that second suggestion doesn't work though it wouldn't bring anything because then you need space to separate expression from
if.$endgroup$
– Nishioka
Jan 6 at 14:34
$begingroup$
True ... alternative 149.
$endgroup$
– Jonathan Frech
Jan 6 at 15:03
$begingroup$
True ... alternative 149.
$endgroup$
– Jonathan Frech
Jan 6 at 15:03
add a comment |
$begingroup$
Python 2, 78 74 73 bytes
l=lambda n,a=0,c=0,s="":c*(n==s)or a and l(n,a-1,c+1,s+`a-1`)or l(n,a-~c)
Try it online!
-1 byte thanks to Arnauld
Takes input as a string. The program rather quickly runs into Python's recursion depth limit, but it can finish most of the test cases.
How it works
l=lambda n, # The input number, in the form of a string
a=0, # The program will attempt to reconstruct n by
# building a string by pasting decreasing
# numbers, stored in a, after each other.
c=0, # A counter of the amount of numbers
s="": # The current constructed string
c*(n==s) # Return the counter if s matches n
or # Else
a and l(n,a-1,c+1,s+`a-1`) # If a is not yet zero, paste a-1 after s
or # Else
l(n,a-~c) # Start again, from one higher than last time
answered Jan 14 at 20:23
ArBoArBo
32115
$endgroup$
1
$begingroup$
Nice answer!a+c+1can be shortened toa-~c.
$endgroup$
– Arnauld
Jan 14 at 21:01
add a comment |
$begingroup$
Python 2, 78 74 73 bytes
l=lambda n,a=0,c=0,s="":c*(n==s)or a and l(n,a-1,c+1,s+`a-1`)or l(n,a-~c)
Try it online!
-1 byte thanks to Arnauld
Takes input as a string. The program rather quickly runs into Python's recursion depth limit, but it can finish most of the test cases.
How it works
l=lambda n, # The input number, in the form of a string
a=0, # The program will attempt to reconstruct n by
# building a string by pasting decreasing
# numbers, stored in a, after each other.
c=0, # A counter of the amount of numbers
s="": # The current constructed string
c*(n==s) # Return the counter if s matches n
or # Else
a and l(n,a-1,c+1,s+`a-1`) # If a is not yet zero, paste a-1 after s
or # Else
l(n,a-~c) # Start again, from one higher than last time
answered Jan 14 at 20:23
ArBoArBo
32115
$endgroup$
1
$begingroup$
Nice answer!a+c+1can be shortened toa-~c.
$endgroup$
– Arnauld
Jan 14 at 21:01
add a comment |
$begingroup$
Python 2, 78 74 73 bytes
l=lambda n,a=0,c=0,s="":c*(n==s)or a and l(n,a-1,c+1,s+`a-1`)or l(n,a-~c)
Try it online!
-1 byte thanks to Arnauld
Takes input as a string. The program rather quickly runs into Python's recursion depth limit, but it can finish most of the test cases.
How it works
l=lambda n, # The input number, in the form of a string
a=0, # The program will attempt to reconstruct n by
# building a string by pasting decreasing
# numbers, stored in a, after each other.
c=0, # A counter of the amount of numbers
s="": # The current constructed string
c*(n==s) # Return the counter if s matches n
or # Else
a and l(n,a-1,c+1,s+`a-1`) # If a is not yet zero, paste a-1 after s
or # Else
l(n,a-~c) # Start again, from one higher than last time
answered Jan 14 at 20:23
ArBoArBo
32115
$endgroup$
Python 2, 78 74 73 bytes
l=lambda n,a=0,c=0,s="":c*(n==s)or a and l(n,a-1,c+1,s+`a-1`)or l(n,a-~c)
Try it online!
-1 byte thanks to Arnauld
Takes input as a string. The program rather quickly runs into Python's recursion depth limit, but it can finish most of the test cases.
How it works
l=lambda n, # The input number, in the form of a string
a=0, # The program will attempt to reconstruct n by
# building a string by pasting decreasing
# numbers, stored in a, after each other.
c=0, # A counter of the amount of numbers
s="": # The current constructed string
c*(n==s) # Return the counter if s matches n
or # Else
a and l(n,a-1,c+1,s+`a-1`) # If a is not yet zero, paste a-1 after s
or # Else
l(n,a-~c) # Start again, from one higher than last time
answered Jan 14 at 20:23
ArBoArBo
32115
edited Jan 15 at 14:15
answered Jan 14 at 20:23
ArBoArBo
32115
answered Jan 14 at 20:23
ArBoArBo
32115
answered Jan 14 at 20:23
ArBoArBo
32115
32115
1
$begingroup$
Nice answer!a+c+1can be shortened toa-~c.
$endgroup$
– Arnauld
Jan 14 at 21:01
add a comment |
1
$begingroup$
Nice answer!a+c+1can be shortened toa-~c.
$endgroup$
– Arnauld
Jan 14 at 21:01
1
1
$begingroup$
Nice answer!
a+c+1 can be shortened to a-~c.$endgroup$
– Arnauld
Jan 14 at 21:01
$begingroup$
Nice answer!
a+c+1 can be shortened to a-~c.$endgroup$
– Arnauld
Jan 14 at 21:01
add a comment |
$begingroup$
05AB1E, 10 bytes
ÝRŒʒJQ}€gà
Extremely slow, so the TIO below only works for test cases below 750..
Try it online.
Explanation:
Ý # Create a list in the range [0, (implicit) input]
# i.e. 109 → [0,1,2,...,107,108,109]
R # Reverse it
# i.e. [0,1,2,...,107,108,109] → [109,108,107,...,2,1,0]
Π# Get all possible sublists of this list
# i.e. [109,108,107,...,2,1,0]
# → [[109],[109,108],[109,108,107],...,[2,1,0],[1],[1,0],[0]]
ʒ } # Filter it by:
J # Where the sublist joined together
# i.e. [10,9] → "109"
# i.e. [109,108,107] → "109108107"
Q # Are equal to the (implicit) input
# i.e. 109 and "109" → 1 (truthy)
# i.e. 109 and "109108107" → 0 (falsey)
€g # After filtering, take the length of each remaining inner list
# i.e. [[109],[[10,9]] → [1,2]
à # And only leave the maximum length (which is output implicitly)
# i.e. [1,2] → 2
answered Jan 4 at 12:18
Kevin CruijssenKevin Cruijssen
36.5k555192
$endgroup$
2
$begingroup$
Code golf - where adding 1 byte to your program to go fromn!ton lg njust isn't worth it.
$endgroup$
– corsiKa
Jan 5 at 20:49
add a comment |
$begingroup$
05AB1E, 10 bytes
ÝRŒʒJQ}€gà
Extremely slow, so the TIO below only works for test cases below 750..
Try it online.
Explanation:
Ý # Create a list in the range [0, (implicit) input]
# i.e. 109 → [0,1,2,...,107,108,109]
R # Reverse it
# i.e. [0,1,2,...,107,108,109] → [109,108,107,...,2,1,0]
Π# Get all possible sublists of this list
# i.e. [109,108,107,...,2,1,0]
# → [[109],[109,108],[109,108,107],...,[2,1,0],[1],[1,0],[0]]
ʒ } # Filter it by:
J # Where the sublist joined together
# i.e. [10,9] → "109"
# i.e. [109,108,107] → "109108107"
Q # Are equal to the (implicit) input
# i.e. 109 and "109" → 1 (truthy)
# i.e. 109 and "109108107" → 0 (falsey)
€g # After filtering, take the length of each remaining inner list
# i.e. [[109],[[10,9]] → [1,2]
à # And only leave the maximum length (which is output implicitly)
# i.e. [1,2] → 2
answered Jan 4 at 12:18
Kevin CruijssenKevin Cruijssen
36.5k555192
$endgroup$
2
$begingroup$
Code golf - where adding 1 byte to your program to go fromn!ton lg njust isn't worth it.
$endgroup$
– corsiKa
Jan 5 at 20:49
add a comment |
$begingroup$
05AB1E, 10 bytes
ÝRŒʒJQ}€gà
Extremely slow, so the TIO below only works for test cases below 750..
Try it online.
Explanation:
Ý # Create a list in the range [0, (implicit) input]
# i.e. 109 → [0,1,2,...,107,108,109]
R # Reverse it
# i.e. [0,1,2,...,107,108,109] → [109,108,107,...,2,1,0]
Π# Get all possible sublists of this list
# i.e. [109,108,107,...,2,1,0]
# → [[109],[109,108],[109,108,107],...,[2,1,0],[1],[1,0],[0]]
ʒ } # Filter it by:
J # Where the sublist joined together
# i.e. [10,9] → "109"
# i.e. [109,108,107] → "109108107"
Q # Are equal to the (implicit) input
# i.e. 109 and "109" → 1 (truthy)
# i.e. 109 and "109108107" → 0 (falsey)
€g # After filtering, take the length of each remaining inner list
# i.e. [[109],[[10,9]] → [1,2]
à # And only leave the maximum length (which is output implicitly)
# i.e. [1,2] → 2
answered Jan 4 at 12:18
Kevin CruijssenKevin Cruijssen
36.5k555192
$endgroup$
05AB1E, 10 bytes
ÝRŒʒJQ}€gà
Extremely slow, so the TIO below only works for test cases below 750..
Try it online.
Explanation:
Ý # Create a list in the range [0, (implicit) input]
# i.e. 109 → [0,1,2,...,107,108,109]
R # Reverse it
# i.e. [0,1,2,...,107,108,109] → [109,108,107,...,2,1,0]
Π# Get all possible sublists of this list
# i.e. [109,108,107,...,2,1,0]
# → [[109],[109,108],[109,108,107],...,[2,1,0],[1],[1,0],[0]]
ʒ } # Filter it by:
J # Where the sublist joined together
# i.e. [10,9] → "109"
# i.e. [109,108,107] → "109108107"
Q # Are equal to the (implicit) input
# i.e. 109 and "109" → 1 (truthy)
# i.e. 109 and "109108107" → 0 (falsey)
€g # After filtering, take the length of each remaining inner list
# i.e. [[109],[[10,9]] → [1,2]
à # And only leave the maximum length (which is output implicitly)
# i.e. [1,2] → 2
answered Jan 4 at 12:18
Kevin CruijssenKevin Cruijssen
36.5k555192
edited Jan 4 at 12:28
answered Jan 4 at 12:18
Kevin CruijssenKevin Cruijssen
36.5k555192
answered Jan 4 at 12:18
Kevin CruijssenKevin Cruijssen
36.5k555192
answered Jan 4 at 12:18
Kevin CruijssenKevin Cruijssen
36.5k555192
36.5k555192
2
$begingroup$
Code golf - where adding 1 byte to your program to go fromn!ton lg njust isn't worth it.
$endgroup$
– corsiKa
Jan 5 at 20:49
add a comment |
2
$begingroup$
Code golf - where adding 1 byte to your program to go fromn!ton lg njust isn't worth it.
$endgroup$
– corsiKa
Jan 5 at 20:49
2
2
$begingroup$
Code golf - where adding 1 byte to your program to go from
n! to n lg n just isn't worth it.$endgroup$
– corsiKa
Jan 5 at 20:49
$begingroup$
Code golf - where adding 1 byte to your program to go from
n! to n lg n just isn't worth it.$endgroup$
– corsiKa
Jan 5 at 20:49
add a comment |
$begingroup$
Pyth, 16 bytes
lef!.EhM.+vMT./z
Try it online here, or verify all the test cases at once here.
lef!.EhM.+vMT./z Implicit: z=input as string
./z Get all divisions of z into disjoint substrings
f Filter the above, as T, keeping those where the following is truthy:
vMT Parse each substring as an int
.+ Get difference between each pair
hM Increment each
!.E Are all elements 0? { NOT(ANY(...)) }
e Take the last element of the filtered divisions
Divisions are generated with fewest substrings first, so last remaining division is also the longest
l Length of the above, implicit print
answered Jan 4 at 12:52
SokSok
3,637723
$endgroup$
add a comment |
$begingroup$
Pyth, 16 bytes
lef!.EhM.+vMT./z
Try it online here, or verify all the test cases at once here.
lef!.EhM.+vMT./z Implicit: z=input as string
./z Get all divisions of z into disjoint substrings
f Filter the above, as T, keeping those where the following is truthy:
vMT Parse each substring as an int
.+ Get difference between each pair
hM Increment each
!.E Are all elements 0? { NOT(ANY(...)) }
e Take the last element of the filtered divisions
Divisions are generated with fewest substrings first, so last remaining division is also the longest
l Length of the above, implicit print
answered Jan 4 at 12:52
SokSok
3,637723
$endgroup$
add a comment |
$begingroup$
Pyth, 16 bytes
lef!.EhM.+vMT./z
Try it online here, or verify all the test cases at once here.
lef!.EhM.+vMT./z Implicit: z=input as string
./z Get all divisions of z into disjoint substrings
f Filter the above, as T, keeping those where the following is truthy:
vMT Parse each substring as an int
.+ Get difference between each pair
hM Increment each
!.E Are all elements 0? { NOT(ANY(...)) }
e Take the last element of the filtered divisions
Divisions are generated with fewest substrings first, so last remaining division is also the longest
l Length of the above, implicit print
answered Jan 4 at 12:52
SokSok
3,637723
$endgroup$
Pyth, 16 bytes
lef!.EhM.+vMT./z
Try it online here, or verify all the test cases at once here.
lef!.EhM.+vMT./z Implicit: z=input as string
./z Get all divisions of z into disjoint substrings
f Filter the above, as T, keeping those where the following is truthy:
vMT Parse each substring as an int
.+ Get difference between each pair
hM Increment each
!.E Are all elements 0? { NOT(ANY(...)) }
e Take the last element of the filtered divisions
Divisions are generated with fewest substrings first, so last remaining division is also the longest
l Length of the above, implicit print
answered Jan 4 at 12:52
SokSok
3,637723
answered Jan 4 at 12:52
SokSok
3,637723
answered Jan 4 at 12:52
SokSok
3,637723
answered Jan 4 at 12:52
SokSok
3,637723
3,637723
add a comment |
add a comment |
$begingroup$
Charcoal, 26 bytes
Ｆ⊕ＬθＦ⊕Ｌθ⊞υ⭆κ⁻Ｉ…θιλＩ﹪⌕υθ⊕Ｌθ
Try it online! Link is to verbose version of code. Explanation:
Ｆ⊕Ｌθ
Loop i from 0 to the length of the input.
Ｆ⊕Ｌθ
Loop k from 0 to the length of the input.
⊞υ⭆κ⁻Ｉ…θ⊕ιλ
Calculate the first k numbers in the descending sequence starting from the number given by the first i digits of the input, concatenate them, and accumulate each resulting string in the predefined empty list.
Ｉ﹪⌕υθ⊕Ｌθ
Find the position of the first matching copy of the input and reduce it modulo 1 more than the length of the input.
Example: For an input of 2019 the following strings are generated:
0
1 0
2 0-1
3 0-1-2
4 0-1-2-3
5
6 2
7 21
8 210
9 210-1
10
11 20
12 2019
13 201918
14 20191817
15
16 201
17 201200
18 201200199
19 201200199198
20
21 2019
22 20192018
23 201920182017
24 2019201820172016
2019 is then found at index 12, which is reduced modulo 5 to give 2, the desired answer.
answered Jan 4 at 21:20
NeilNeil
79.9k744178
$endgroup$
add a comment |
$begingroup$
Charcoal, 26 bytes
Ｆ⊕ＬθＦ⊕Ｌθ⊞υ⭆κ⁻Ｉ…θιλＩ﹪⌕υθ⊕Ｌθ
Try it online! Link is to verbose version of code. Explanation:
Ｆ⊕Ｌθ
Loop i from 0 to the length of the input.
Ｆ⊕Ｌθ
Loop k from 0 to the length of the input.
⊞υ⭆κ⁻Ｉ…θ⊕ιλ
Calculate the first k numbers in the descending sequence starting from the number given by the first i digits of the input, concatenate them, and accumulate each resulting string in the predefined empty list.
Ｉ﹪⌕υθ⊕Ｌθ
Find the position of the first matching copy of the input and reduce it modulo 1 more than the length of the input.
Example: For an input of 2019 the following strings are generated:
0
1 0
2 0-1
3 0-1-2
4 0-1-2-3
5
6 2
7 21
8 210
9 210-1
10
11 20
12 2019
13 201918
14 20191817
15
16 201
17 201200
18 201200199
19 201200199198
20
21 2019
22 20192018
23 201920182017
24 2019201820172016
2019 is then found at index 12, which is reduced modulo 5 to give 2, the desired answer.
answered Jan 4 at 21:20
NeilNeil
79.9k744178
$endgroup$
add a comment |
$begingroup$
Charcoal, 26 bytes
Ｆ⊕ＬθＦ⊕Ｌθ⊞υ⭆κ⁻Ｉ…θιλＩ﹪⌕υθ⊕Ｌθ
Try it online! Link is to verbose version of code. Explanation:
Ｆ⊕Ｌθ
Loop i from 0 to the length of the input.
Ｆ⊕Ｌθ
Loop k from 0 to the length of the input.
⊞υ⭆κ⁻Ｉ…θ⊕ιλ
Calculate the first k numbers in the descending sequence starting from the number given by the first i digits of the input, concatenate them, and accumulate each resulting string in the predefined empty list.
Ｉ﹪⌕υθ⊕Ｌθ
Find the position of the first matching copy of the input and reduce it modulo 1 more than the length of the input.
Example: For an input of 2019 the following strings are generated:
0
1 0
2 0-1
3 0-1-2
4 0-1-2-3
5
6 2
7 21
8 210
9 210-1
10
11 20
12 2019
13 201918
14 20191817
15
16 201
17 201200
18 201200199
19 201200199198
20
21 2019
22 20192018
23 201920182017
24 2019201820172016
2019 is then found at index 12, which is reduced modulo 5 to give 2, the desired answer.
answered Jan 4 at 21:20
NeilNeil
79.9k744178
$endgroup$
Charcoal, 26 bytes
Ｆ⊕ＬθＦ⊕Ｌθ⊞υ⭆κ⁻Ｉ…θιλＩ﹪⌕υθ⊕Ｌθ
Try it online! Link is to verbose version of code. Explanation:
Ｆ⊕Ｌθ
Loop i from 0 to the length of the input.
Ｆ⊕Ｌθ
Loop k from 0 to the length of the input.
⊞υ⭆κ⁻Ｉ…θ⊕ιλ
Calculate the first k numbers in the descending sequence starting from the number given by the first i digits of the input, concatenate them, and accumulate each resulting string in the predefined empty list.
Ｉ﹪⌕υθ⊕Ｌθ
Find the position of the first matching copy of the input and reduce it modulo 1 more than the length of the input.
Example: For an input of 2019 the following strings are generated:
0
1 0
2 0-1
3 0-1-2
4 0-1-2-3
5
6 2
7 21
8 210
9 210-1
10
11 20
12 2019
13 201918
14 20191817
15
16 201
17 201200
18 201200199
19 201200199198
20
21 2019
22 20192018
23 201920182017
24 2019201820172016
2019 is then found at index 12, which is reduced modulo 5 to give 2, the desired answer.
answered Jan 4 at 21:20
NeilNeil
79.9k744178
answered Jan 4 at 21:20
NeilNeil
79.9k744178
answered Jan 4 at 21:20
NeilNeil
79.9k744178
answered Jan 4 at 21:20
NeilNeil
79.9k744178
79.9k744178
add a comment |
add a comment |
$begingroup$
Haskell, 87 bytes
maximum.map length.(0#)
a#(b:c)=[a:x|c==||b>0,x<-b#c,a==x!!0+1]++(10*a+b)#c
a#b=[[a]]
Input is a list of digits.
Try it online!
Function # builds a list of all possible splits by looking at both
- prepending the current number
ato all splits returned by a recursive call with the rest of the input (x<-b#c), but only if the next number not zero (b>0) (or it's the last number in the input (c==)) andais one greater than the first number of the respective previous splitx(a==x!!0+1).
and
- appending the next digit
bfrom the input list to the current numberaand going on with the rest of the input ((10*a+b)#c)
Base case is when the input list is empty (i.e. does not pattern match (b:c)). The recursion starts with the current number a being 0 ((0#)), which never hits the first branch (prepending a to all previous splits), because it will never be greater than any number of the splits.
Take the length of each split and find the maximum (maximum.map length).
A variant with also 87 bytes:
fst.maximum.(0#)
a#(b:c)=[(r+1,a)|c==||b>0,(r,x)<-b#c,a==x+1]++(10*a+b)#c
a#b=[(1,a)]
which basically works the same way, but instead of keeping the whole split in a list, it only keeps a pair (r,x) of the length of the split r an the first number in the split x.
answered Jan 5 at 15:00
niminimi
31.5k32185
$endgroup$
add a comment |
$begingroup$
Haskell, 87 bytes
maximum.map length.(0#)
a#(b:c)=[a:x|c==||b>0,x<-b#c,a==x!!0+1]++(10*a+b)#c
a#b=[[a]]
Input is a list of digits.
Try it online!
Function # builds a list of all possible splits by looking at both
- prepending the current number
ato all splits returned by a recursive call with the rest of the input (x<-b#c), but only if the next number not zero (b>0) (or it's the last number in the input (c==)) andais one greater than the first number of the respective previous splitx(a==x!!0+1).
and
- appending the next digit
bfrom the input list to the current numberaand going on with the rest of the input ((10*a+b)#c)
Base case is when the input list is empty (i.e. does not pattern match (b:c)). The recursion starts with the current number a being 0 ((0#)), which never hits the first branch (prepending a to all previous splits), because it will never be greater than any number of the splits.
Take the length of each split and find the maximum (maximum.map length).
A variant with also 87 bytes:
fst.maximum.(0#)
a#(b:c)=[(r+1,a)|c==||b>0,(r,x)<-b#c,a==x+1]++(10*a+b)#c
a#b=[(1,a)]
which basically works the same way, but instead of keeping the whole split in a list, it only keeps a pair (r,x) of the length of the split r an the first number in the split x.
answered Jan 5 at 15:00
niminimi
31.5k32185
$endgroup$
add a comment |
$begingroup$
Haskell, 87 bytes
maximum.map length.(0#)
a#(b:c)=[a:x|c==||b>0,x<-b#c,a==x!!0+1]++(10*a+b)#c
a#b=[[a]]
Input is a list of digits.
Try it online!
Function # builds a list of all possible splits by looking at both
- prepending the current number
ato all splits returned by a recursive call with the rest of the input (x<-b#c), but only if the next number not zero (b>0) (or it's the last number in the input (c==)) andais one greater than the first number of the respective previous splitx(a==x!!0+1).
and
- appending the next digit
bfrom the input list to the current numberaand going on with the rest of the input ((10*a+b)#c)
Base case is when the input list is empty (i.e. does not pattern match (b:c)). The recursion starts with the current number a being 0 ((0#)), which never hits the first branch (prepending a to all previous splits), because it will never be greater than any number of the splits.
Take the length of each split and find the maximum (maximum.map length).
A variant with also 87 bytes:
fst.maximum.(0#)
a#(b:c)=[(r+1,a)|c==||b>0,(r,x)<-b#c,a==x+1]++(10*a+b)#c
a#b=[(1,a)]
which basically works the same way, but instead of keeping the whole split in a list, it only keeps a pair (r,x) of the length of the split r an the first number in the split x.
answered Jan 5 at 15:00
niminimi
31.5k32185
$endgroup$
Haskell, 87 bytes
maximum.map length.(0#)
a#(b:c)=[a:x|c==||b>0,x<-b#c,a==x!!0+1]++(10*a+b)#c
a#b=[[a]]
Input is a list of digits.
Try it online!
Function # builds a list of all possible splits by looking at both
- prepending the current number
ato all splits returned by a recursive call with the rest of the input (x<-b#c), but only if the next number not zero (b>0) (or it's the last number in the input (c==)) andais one greater than the first number of the respective previous splitx(a==x!!0+1).
and
- appending the next digit
bfrom the input list to the current numberaand going on with the rest of the input ((10*a+b)#c)
Base case is when the input list is empty (i.e. does not pattern match (b:c)). The recursion starts with the current number a being 0 ((0#)), which never hits the first branch (prepending a to all previous splits), because it will never be greater than any number of the splits.
Take the length of each split and find the maximum (maximum.map length).
A variant with also 87 bytes:
fst.maximum.(0#)
a#(b:c)=[(r+1,a)|c==||b>0,(r,x)<-b#c,a==x+1]++(10*a+b)#c
a#b=[(1,a)]
which basically works the same way, but instead of keeping the whole split in a list, it only keeps a pair (r,x) of the length of the split r an the first number in the split x.
answered Jan 5 at 15:00
niminimi
31.5k32185
edited Jan 5 at 15:44
answered Jan 5 at 15:00
niminimi
31.5k32185
answered Jan 5 at 15:00
niminimi
31.5k32185
answered Jan 5 at 15:00
niminimi
31.5k32185
31.5k32185
add a comment |
add a comment |
$begingroup$
Python 3, 302 282 271 bytes
-10 bytes thanks to the tip by @ElPedro.
Takes input as a string. Basically, it takes increasing larger slices of the number from the left, and sees if for that slice of the number a sequence can be formed using all the numbers.
R=range
I=int
L=len
def g(n,m,t=1):
for i in R(1,L(m)+1):
if I(m)==I(n[:i])+1:
if i==L(n):return-~t
return g(n[i:],n[:i],t+1)
return 1
def f(n):
for i in R(L(n)):
x=n[:i]
for j in R(1,L(x)+1):
if (I(x)==I(n[i:i+j])+1)*I(n[i]):return g(n[i:],x)
return 1
Try it online!
answered Jan 4 at 20:39
Neil A.Neil A.
1,218120
$endgroup$
1
$begingroup$
Since you are usingrange3 times you can defineR=rangeoutside of both functions and then useR(whatever)instead ofrange(whatever)to save 4 bytes.
$endgroup$
– ElPedro
Jan 5 at 17:48
add a comment |
$begingroup$
Python 3, 302 282 271 bytes
-10 bytes thanks to the tip by @ElPedro.
Takes input as a string. Basically, it takes increasing larger slices of the number from the left, and sees if for that slice of the number a sequence can be formed using all the numbers.
R=range
I=int
L=len
def g(n,m,t=1):
for i in R(1,L(m)+1):
if I(m)==I(n[:i])+1:
if i==L(n):return-~t
return g(n[i:],n[:i],t+1)
return 1
def f(n):
for i in R(L(n)):
x=n[:i]
for j in R(1,L(x)+1):
if (I(x)==I(n[i:i+j])+1)*I(n[i]):return g(n[i:],x)
return 1
Try it online!
answered Jan 4 at 20:39
Neil A.Neil A.
1,218120
$endgroup$
1
$begingroup$
Since you are usingrange3 times you can defineR=rangeoutside of both functions and then useR(whatever)instead ofrange(whatever)to save 4 bytes.
$endgroup$
– ElPedro
Jan 5 at 17:48
add a comment |
$begingroup$
Python 3, 302 282 271 bytes
-10 bytes thanks to the tip by @ElPedro.
Takes input as a string. Basically, it takes increasing larger slices of the number from the left, and sees if for that slice of the number a sequence can be formed using all the numbers.
R=range
I=int
L=len
def g(n,m,t=1):
for i in R(1,L(m)+1):
if I(m)==I(n[:i])+1:
if i==L(n):return-~t
return g(n[i:],n[:i],t+1)
return 1
def f(n):
for i in R(L(n)):
x=n[:i]
for j in R(1,L(x)+1):
if (I(x)==I(n[i:i+j])+1)*I(n[i]):return g(n[i:],x)
return 1
Try it online!
answered Jan 4 at 20:39
Neil A.Neil A.
1,218120
$endgroup$
Python 3, 302 282 271 bytes
-10 bytes thanks to the tip by @ElPedro.
Takes input as a string. Basically, it takes increasing larger slices of the number from the left, and sees if for that slice of the number a sequence can be formed using all the numbers.
R=range
I=int
L=len
def g(n,m,t=1):
for i in R(1,L(m)+1):
if I(m)==I(n[:i])+1:
if i==L(n):return-~t
return g(n[i:],n[:i],t+1)
return 1
def f(n):
for i in R(L(n)):
x=n[:i]
for j in R(1,L(x)+1):
if (I(x)==I(n[i:i+j])+1)*I(n[i]):return g(n[i:],x)
return 1
Try it online!
answered Jan 4 at 20:39
Neil A.Neil A.
1,218120
edited Jan 5 at 19:57
answered Jan 4 at 20:39
Neil A.Neil A.
1,218120
answered Jan 4 at 20:39
Neil A.Neil A.
1,218120
answered Jan 4 at 20:39
Neil A.Neil A.
1,218120
1,218120
1
$begingroup$
Since you are usingrange3 times you can defineR=rangeoutside of both functions and then useR(whatever)instead ofrange(whatever)to save 4 bytes.
$endgroup$
– ElPedro
Jan 5 at 17:48
add a comment |
1
$begingroup$
Since you are usingrange3 times you can defineR=rangeoutside of both functions and then useR(whatever)instead ofrange(whatever)to save 4 bytes.
$endgroup$
– ElPedro
Jan 5 at 17:48
1
1
$begingroup$
Since you are using
range 3 times you can define R=range outside of both functions and then use R(whatever) instead of range(whatever) to save 4 bytes.$endgroup$
– ElPedro
Jan 5 at 17:48
$begingroup$
Since you are using
range 3 times you can define R=range outside of both functions and then use R(whatever) instead of range(whatever) to save 4 bytes.$endgroup$
– ElPedro
Jan 5 at 17:48
add a comment |
$begingroup$
Japt, 27 bytes
ò pÊÔpÊqÊfl²i1Uì q"l?"¹ÌèÊÉ
Try it online! or Check most test cases
This doesn't score well, but it uses a unique method and there might be room to golf it a lot more. It also performs well enough that all test cases other than 201200199198 avoid timing out.
Explanation:
ò #Get the range [0...input]
pÊ #Add an "l" to the end
Ô #Reverse it
pÊ #Add an "l" to the end
qÊ #Add an "l" between each number and turn to a string
f ¹ #Find the substrings that match this regex:
l² # The string "ll"
i1 # With this inserted between the "l"s:
Uì # All the digits of the input
q"l?" # With optional spaces between each one
Ì #Get the last match
èÊ #Count the number of "l"s
É #Subtract 1
answered Jan 4 at 20:02
Kamil DrakariKamil Drakari
3,131416
$endgroup$
$begingroup$
I think this works for 27.
$endgroup$
– Shaggy
Jan 5 at 10:30
$begingroup$
25 bytes
$endgroup$
– Shaggy
Jan 5 at 10:43
$begingroup$
@Shaggy both of those fail on input21201because they don't enforce that the sequence end aligns correctly (from my original version the "ends with a comma" line). This or this alternative works.
$endgroup$
– Kamil Drakari
Jan 5 at 15:38
$begingroup$
Ah, OK. In that case: 26 bytes
$endgroup$
– Shaggy
Jan 5 at 15:46
$begingroup$
@Shaggy That and the 28 byte solutions I had fail on210because there isn't a delimiter after 0. Here's a fixed 28 byte that works.
$endgroup$
– Kamil Drakari
Jan 6 at 1:35
|
show 1 more comment
$begingroup$
Japt, 27 bytes
ò pÊÔpÊqÊfl²i1Uì q"l?"¹ÌèÊÉ
Try it online! or Check most test cases
This doesn't score well, but it uses a unique method and there might be room to golf it a lot more. It also performs well enough that all test cases other than 201200199198 avoid timing out.
Explanation:
ò #Get the range [0...input]
pÊ #Add an "l" to the end
Ô #Reverse it
pÊ #Add an "l" to the end
qÊ #Add an "l" between each number and turn to a string
f ¹ #Find the substrings that match this regex:
l² # The string "ll"
i1 # With this inserted between the "l"s:
Uì # All the digits of the input
q"l?" # With optional spaces between each one
Ì #Get the last match
èÊ #Count the number of "l"s
É #Subtract 1
answered Jan 4 at 20:02
Kamil DrakariKamil Drakari
3,131416
$endgroup$
$begingroup$
I think this works for 27.
$endgroup$
– Shaggy
Jan 5 at 10:30
$begingroup$
25 bytes
$endgroup$
– Shaggy
Jan 5 at 10:43
$begingroup$
@Shaggy both of those fail on input21201because they don't enforce that the sequence end aligns correctly (from my original version the "ends with a comma" line). This or this alternative works.
$endgroup$
– Kamil Drakari
Jan 5 at 15:38
$begingroup$
Ah, OK. In that case: 26 bytes
$endgroup$
– Shaggy
Jan 5 at 15:46
$begingroup$
@Shaggy That and the 28 byte solutions I had fail on210because there isn't a delimiter after 0. Here's a fixed 28 byte that works.
$endgroup$
– Kamil Drakari
Jan 6 at 1:35
|
show 1 more comment
$begingroup$
Japt, 27 bytes
ò pÊÔpÊqÊfl²i1Uì q"l?"¹ÌèÊÉ
Try it online! or Check most test cases
This doesn't score well, but it uses a unique method and there might be room to golf it a lot more. It also performs well enough that all test cases other than 201200199198 avoid timing out.
Explanation:
ò #Get the range [0...input]
pÊ #Add an "l" to the end
Ô #Reverse it
pÊ #Add an "l" to the end
qÊ #Add an "l" between each number and turn to a string
f ¹ #Find the substrings that match this regex:
l² # The string "ll"
i1 # With this inserted between the "l"s:
Uì # All the digits of the input
q"l?" # With optional spaces between each one
Ì #Get the last match
èÊ #Count the number of "l"s
É #Subtract 1
answered Jan 4 at 20:02
Kamil DrakariKamil Drakari
3,131416
$endgroup$
Japt, 27 bytes
ò pÊÔpÊqÊfl²i1Uì q"l?"¹ÌèÊÉ
Try it online! or Check most test cases
This doesn't score well, but it uses a unique method and there might be room to golf it a lot more. It also performs well enough that all test cases other than 201200199198 avoid timing out.
Explanation:
ò #Get the range [0...input]
pÊ #Add an "l" to the end
Ô #Reverse it
pÊ #Add an "l" to the end
qÊ #Add an "l" between each number and turn to a string
f ¹ #Find the substrings that match this regex:
l² # The string "ll"
i1 # With this inserted between the "l"s:
Uì # All the digits of the input
q"l?" # With optional spaces between each one
Ì #Get the last match
èÊ #Count the number of "l"s
É #Subtract 1
answered Jan 4 at 20:02
Kamil DrakariKamil Drakari
3,131416
edited Jan 6 at 2:00
answered Jan 4 at 20:02
Kamil DrakariKamil Drakari
3,131416
answered Jan 4 at 20:02
Kamil DrakariKamil Drakari
3,131416
answered Jan 4 at 20:02
Kamil DrakariKamil Drakari
3,131416
3,131416
$begingroup$
I think this works for 27.
$endgroup$
– Shaggy
Jan 5 at 10:30
$begingroup$
25 bytes
$endgroup$
– Shaggy
Jan 5 at 10:43
$begingroup$
@Shaggy both of those fail on input21201because they don't enforce that the sequence end aligns correctly (from my original version the "ends with a comma" line). This or this alternative works.
$endgroup$
– Kamil Drakari
Jan 5 at 15:38
$begingroup$
Ah, OK. In that case: 26 bytes
$endgroup$
– Shaggy
Jan 5 at 15:46
$begingroup$
@Shaggy That and the 28 byte solutions I had fail on210because there isn't a delimiter after 0. Here's a fixed 28 byte that works.
$endgroup$
– Kamil Drakari
Jan 6 at 1:35
|
show 1 more comment
$begingroup$
I think this works for 27.
$endgroup$
– Shaggy
Jan 5 at 10:30
$begingroup$
25 bytes
$endgroup$
– Shaggy
Jan 5 at 10:43
$begingroup$
@Shaggy both of those fail on input21201because they don't enforce that the sequence end aligns correctly (from my original version the "ends with a comma" line). This or this alternative works.
$endgroup$
– Kamil Drakari
Jan 5 at 15:38
$begingroup$
Ah, OK. In that case: 26 bytes
$endgroup$
– Shaggy
Jan 5 at 15:46
$begingroup$
@Shaggy That and the 28 byte solutions I had fail on210because there isn't a delimiter after 0. Here's a fixed 28 byte that works.
$endgroup$
– Kamil Drakari
Jan 6 at 1:35
$begingroup$
I think this works for 27.
$endgroup$
– Shaggy
Jan 5 at 10:30
$begingroup$
I think this works for 27.
$endgroup$
– Shaggy
Jan 5 at 10:30
$begingroup$
25 bytes
$endgroup$
– Shaggy
Jan 5 at 10:43
$begingroup$
25 bytes
$endgroup$
– Shaggy
Jan 5 at 10:43
$begingroup$
@Shaggy both of those fail on input
21201 because they don't enforce that the sequence end aligns correctly (from my original version the "ends with a comma" line). This or this alternative works.$endgroup$
– Kamil Drakari
Jan 5 at 15:38
$begingroup$
@Shaggy both of those fail on input
21201 because they don't enforce that the sequence end aligns correctly (from my original version the "ends with a comma" line). This or this alternative works.$endgroup$
– Kamil Drakari
Jan 5 at 15:38
$begingroup$
Ah, OK. In that case: 26 bytes
$endgroup$
– Shaggy
Jan 5 at 15:46
$begingroup$
Ah, OK. In that case: 26 bytes
$endgroup$
– Shaggy
Jan 5 at 15:46
$begingroup$
@Shaggy That and the 28 byte solutions I had fail on
210 because there isn't a delimiter after 0. Here's a fixed 28 byte that works.$endgroup$
– Kamil Drakari
Jan 6 at 1:35
$begingroup$
@Shaggy That and the 28 byte solutions I had fail on
210 because there isn't a delimiter after 0. Here's a fixed 28 byte that works.$endgroup$
– Kamil Drakari
Jan 6 at 1:35
|
show 1 more comment
$begingroup$
Jelly, 11 bytes
ŒṖḌ’DɗƑƇẈṀ
Byte for byte, no match for the other Jelly solution, but this one should be roughly $Oleft(n^{0.3}right)$.
Try it online!
How it works
ŒṖḌ’DɗƑƇẈṀ Main link. Argument: n (integer)
ŒṖ Yield all partitions of n's digit list in base 10.
Ƈ Comb; keep only partitions for which the link to the left returns 1.
Ƒ Fixed; yield 1 if calling the link to the left returns its argument.
Cumulatively reduce the partition by the link to the left.
ɗ Combine the three links to the left into a dyadic chain.
Ḍ Undecimal; convert a digit list into an integer.
’ Decrement the result.
D Decimal; convert the integer back to a digit list.
answered Jan 4 at 14:02
Dennis♦Dennis
187k32297737
$endgroup$
add a comment |
$begingroup$
Jelly, 11 bytes
ŒṖḌ’DɗƑƇẈṀ
Byte for byte, no match for the other Jelly solution, but this one should be roughly $Oleft(n^{0.3}right)$.
Try it online!
How it works
ŒṖḌ’DɗƑƇẈṀ Main link. Argument: n (integer)
ŒṖ Yield all partitions of n's digit list in base 10.
Ƈ Comb; keep only partitions for which the link to the left returns 1.
Ƒ Fixed; yield 1 if calling the link to the left returns its argument.
Cumulatively reduce the partition by the link to the left.
ɗ Combine the three links to the left into a dyadic chain.
Ḍ Undecimal; convert a digit list into an integer.
’ Decrement the result.
D Decimal; convert the integer back to a digit list.
answered Jan 4 at 14:02
Dennis♦Dennis
187k32297737
$endgroup$
add a comment |
$begingroup$
Jelly, 11 bytes
ŒṖḌ’DɗƑƇẈṀ
Byte for byte, no match for the other Jelly solution, but this one should be roughly $Oleft(n^{0.3}right)$.
Try it online!
How it works
ŒṖḌ’DɗƑƇẈṀ Main link. Argument: n (integer)
ŒṖ Yield all partitions of n's digit list in base 10.
Ƈ Comb; keep only partitions for which the link to the left returns 1.
Ƒ Fixed; yield 1 if calling the link to the left returns its argument.
Cumulatively reduce the partition by the link to the left.
ɗ Combine the three links to the left into a dyadic chain.
Ḍ Undecimal; convert a digit list into an integer.
’ Decrement the result.
D Decimal; convert the integer back to a digit list.
answered Jan 4 at 14:02
Dennis♦Dennis
187k32297737
$endgroup$
Jelly, 11 bytes
ŒṖḌ’DɗƑƇẈṀ
Byte for byte, no match for the other Jelly solution, but this one should be roughly $Oleft(n^{0.3}right)$.
Try it online!
How it works
ŒṖḌ’DɗƑƇẈṀ Main link. Argument: n (integer)
ŒṖ Yield all partitions of n's digit list in base 10.
Ƈ Comb; keep only partitions for which the link to the left returns 1.
Ƒ Fixed; yield 1 if calling the link to the left returns its argument.
Cumulatively reduce the partition by the link to the left.
ɗ Combine the three links to the left into a dyadic chain.
Ḍ Undecimal; convert a digit list into an integer.
’ Decrement the result.
D Decimal; convert the integer back to a digit list.
answered Jan 4 at 14:02
Dennis♦Dennis
187k32297737
edited Jan 4 at 20:29
answered Jan 4 at 14:02
Dennis♦Dennis
187k32297737
answered Jan 4 at 14:02
Dennis♦Dennis
187k32297737
answered Jan 4 at 14:02
Dennis♦Dennis
187k32297737
187k32297737
add a comment |
add a comment |
$begingroup$
Haskell, 65 bytes
f i=[y|x<-[0..],y<-[1..length i],i==(show=<<[x+y-1,x+y-2..x])]!!0
Input is a string.
Try it online!
Completely different to my other answer. A simple brute force that tries all lists of consecutive descending numbers until it finds one that equals the input list.
If we limit the input number to 64-bit integers, we can save 6 bytes by looping y through [1..19], because the largest 64-bit integer has 19 digits and there's no need to test lists with more elements.
Haskell, 59 bytes
f i=[y|x<-[0..],y<-[1..19],i==(show=<<[x+y-1,x+y-2..x])]!!0
Try it online!
answered Jan 5 at 16:02
niminimi
31.5k32185
$endgroup$
add a comment |
$begingroup$
Haskell, 65 bytes
f i=[y|x<-[0..],y<-[1..length i],i==(show=<<[x+y-1,x+y-2..x])]!!0
Input is a string.
Try it online!
Completely different to my other answer. A simple brute force that tries all lists of consecutive descending numbers until it finds one that equals the input list.
If we limit the input number to 64-bit integers, we can save 6 bytes by looping y through [1..19], because the largest 64-bit integer has 19 digits and there's no need to test lists with more elements.
Haskell, 59 bytes
f i=[y|x<-[0..],y<-[1..19],i==(show=<<[x+y-1,x+y-2..x])]!!0
Try it online!
answered Jan 5 at 16:02
niminimi
31.5k32185
$endgroup$
add a comment |
$begingroup$
Haskell, 65 bytes
f i=[y|x<-[0..],y<-[1..length i],i==(show=<<[x+y-1,x+y-2..x])]!!0
Input is a string.
Try it online!
Completely different to my other answer. A simple brute force that tries all lists of consecutive descending numbers until it finds one that equals the input list.
If we limit the input number to 64-bit integers, we can save 6 bytes by looping y through [1..19], because the largest 64-bit integer has 19 digits and there's no need to test lists with more elements.
Haskell, 59 bytes
f i=[y|x<-[0..],y<-[1..19],i==(show=<<[x+y-1,x+y-2..x])]!!0
Try it online!
answered Jan 5 at 16:02
niminimi
31.5k32185
$endgroup$
Haskell, 65 bytes
f i=[y|x<-[0..],y<-[1..length i],i==(show=<<[x+y-1,x+y-2..x])]!!0
Input is a string.
Try it online!
Completely different to my other answer. A simple brute force that tries all lists of consecutive descending numbers until it finds one that equals the input list.
If we limit the input number to 64-bit integers, we can save 6 bytes by looping y through [1..19], because the largest 64-bit integer has 19 digits and there's no need to test lists with more elements.
Haskell, 59 bytes
f i=[y|x<-[0..],y<-[1..19],i==(show=<<[x+y-1,x+y-2..x])]!!0
Try it online!
answered Jan 5 at 16:02
niminimi
31.5k32185
edited Jan 5 at 16:47
answered Jan 5 at 16:02
niminimi
31.5k32185
answered Jan 5 at 16:02
niminimi
31.5k32185
answered Jan 5 at 16:02
niminimi
31.5k32185
31.5k32185
add a comment |
add a comment |
$begingroup$
Python 2, 95 bytes
lambda n:max(j-i for j in range(n+1)for i in range(-1,j)if''.join(map(str,range(j,i,-1)))==`n`)
Another slow, brute-force solution.
Try it online!
answered Jan 5 at 20:32
Dennis♦Dennis
187k32297737
$endgroup$
add a comment |
$begingroup$
Python 2, 95 bytes
lambda n:max(j-i for j in range(n+1)for i in range(-1,j)if''.join(map(str,range(j,i,-1)))==`n`)
Another slow, brute-force solution.
Try it online!
answered Jan 5 at 20:32
Dennis♦Dennis
187k32297737
$endgroup$
add a comment |
$begingroup$
Python 2, 95 bytes
lambda n:max(j-i for j in range(n+1)for i in range(-1,j)if''.join(map(str,range(j,i,-1)))==`n`)
Another slow, brute-force solution.
Try it online!
answered Jan 5 at 20:32
Dennis♦Dennis
187k32297737
$endgroup$
Python 2, 95 bytes
lambda n:max(j-i for j in range(n+1)for i in range(-1,j)if''.join(map(str,range(j,i,-1)))==`n`)
Another slow, brute-force solution.
Try it online!
answered Jan 5 at 20:32
Dennis♦Dennis
187k32297737
answered Jan 5 at 20:32
Dennis♦Dennis
187k32297737
answered Jan 5 at 20:32
Dennis♦Dennis
187k32297737
answered Jan 5 at 20:32
Dennis♦Dennis
187k32297737
187k32297737
add a comment |
add a comment |
$begingroup$
Dyalog APL, 138 bytes
Bit of a mouthful, but it works fast for big numbers too. If you try it online, prefix the dfn by ⎕← and provide input on the right as a list of digits.
{⌈/⍵((≢⊂)×1∧.=2-/10⊥¨⊂)⍨⍤1⊢1,{⍬≡1↓⍵:↑⍬1⋄0=⊃⍵:0,∇1↓⍵⋄↑,0 1∘.,⊂⍤1∇1↓⍵}1↓⍵}
Explanation
First, the inner dfn on the right which recursively constructs a list of possible ways to partition (with ⊂) the list of digits. For example 1 0 1 0 ⊂ 2 0 1 9 returns the nested vector (2 0)(1 9).
{
⍬≡1↓⍵: ↑⍬1 ⍝ Edge case: If ⍵ is singleton list, return the column matrix (0 1)
0=⊃⍵: 0,∇1↓⍵ ⍝ If head of ⍵ is 0, return 0 catenated to this dfn called on tail ⍵
↑,0 1∘.,⊂⍤1∇1↓⍵ ⍝ Finds 1 cat recursive call on tail ⍵ and 0 cat recursive call on ⍵.
} ⍝ Makes a matrix with a row for each possibility.
We use 1, to add a column of 1s at the start and end up with a matrix of valid partitions for ⍵.
Now the function train on the left in parens. Due to ⍨ the left argument to the train is the row of the partitions matrix and the right argument is the user input.
The train is a pile of forks with an atop as the far left tine.
((≢⊂)×1∧.=2-/10⊥¨⊂)⍨ ⍝ ⍨ swaps left and right arguments of the train.
⊂ ⍝ Partition ⍵ according to ⍺.
10⊥¨ ⍝ Decode each partition (turns strings of digits into numbers)
2-/ ⍝ Difference between adjacent cells
1∧.= ⍝ All equal 1?
⊂ ⍝ Partition ⍵ according to ⍺ again
≢ ⍝ Number of cells (ie number of partitions)
× ⍝ Multiply.
If the partition creates a sequence of descending numbers, the train returns the length of the sequence. Otherwise zero.
⍤1⊢ applies the function train between the user input and each row of the partitions matrix, returning a value for each row of the matrix. ⊢ is necessary to disambiguate between operand to ⍤ and argument to the derived function of ⍤.
⌈/ finds the maximum.
Could find a shorter algorithm but I wanted to try this way which is the most direct and declarative I could think of.
answered Jan 12 at 21:39
akhmornakhmorn
312
$endgroup$
$begingroup$
Welcome to PPCG! This is an impressive first post!
$endgroup$
– Riker
Jan 12 at 21:47
add a comment |
$begingroup$
Dyalog APL, 138 bytes
Bit of a mouthful, but it works fast for big numbers too. If you try it online, prefix the dfn by ⎕← and provide input on the right as a list of digits.
{⌈/⍵((≢⊂)×1∧.=2-/10⊥¨⊂)⍨⍤1⊢1,{⍬≡1↓⍵:↑⍬1⋄0=⊃⍵:0,∇1↓⍵⋄↑,0 1∘.,⊂⍤1∇1↓⍵}1↓⍵}
Explanation
First, the inner dfn on the right which recursively constructs a list of possible ways to partition (with ⊂) the list of digits. For example 1 0 1 0 ⊂ 2 0 1 9 returns the nested vector (2 0)(1 9).
{
⍬≡1↓⍵: ↑⍬1 ⍝ Edge case: If ⍵ is singleton list, return the column matrix (0 1)
0=⊃⍵: 0,∇1↓⍵ ⍝ If head of ⍵ is 0, return 0 catenated to this dfn called on tail ⍵
↑,0 1∘.,⊂⍤1∇1↓⍵ ⍝ Finds 1 cat recursive call on tail ⍵ and 0 cat recursive call on ⍵.
} ⍝ Makes a matrix with a row for each possibility.
We use 1, to add a column of 1s at the start and end up with a matrix of valid partitions for ⍵.
Now the function train on the left in parens. Due to ⍨ the left argument to the train is the row of the partitions matrix and the right argument is the user input.
The train is a pile of forks with an atop as the far left tine.
((≢⊂)×1∧.=2-/10⊥¨⊂)⍨ ⍝ ⍨ swaps left and right arguments of the train.
⊂ ⍝ Partition ⍵ according to ⍺.
10⊥¨ ⍝ Decode each partition (turns strings of digits into numbers)
2-/ ⍝ Difference between adjacent cells
1∧.= ⍝ All equal 1?
⊂ ⍝ Partition ⍵ according to ⍺ again
≢ ⍝ Number of cells (ie number of partitions)
× ⍝ Multiply.
If the partition creates a sequence of descending numbers, the train returns the length of the sequence. Otherwise zero.
⍤1⊢ applies the function train between the user input and each row of the partitions matrix, returning a value for each row of the matrix. ⊢ is necessary to disambiguate between operand to ⍤ and argument to the derived function of ⍤.
⌈/ finds the maximum.
Could find a shorter algorithm but I wanted to try this way which is the most direct and declarative I could think of.
answered Jan 12 at 21:39
akhmornakhmorn
312
$endgroup$
$begingroup$
Welcome to PPCG! This is an impressive first post!
$endgroup$
– Riker
Jan 12 at 21:47
add a comment |
$begingroup$
Dyalog APL, 138 bytes
Bit of a mouthful, but it works fast for big numbers too. If you try it online, prefix the dfn by ⎕← and provide input on the right as a list of digits.
{⌈/⍵((≢⊂)×1∧.=2-/10⊥¨⊂)⍨⍤1⊢1,{⍬≡1↓⍵:↑⍬1⋄0=⊃⍵:0,∇1↓⍵⋄↑,0 1∘.,⊂⍤1∇1↓⍵}1↓⍵}
Explanation
First, the inner dfn on the right which recursively constructs a list of possible ways to partition (with ⊂) the list of digits. For example 1 0 1 0 ⊂ 2 0 1 9 returns the nested vector (2 0)(1 9).
{
⍬≡1↓⍵: ↑⍬1 ⍝ Edge case: If ⍵ is singleton list, return the column matrix (0 1)
0=⊃⍵: 0,∇1↓⍵ ⍝ If head of ⍵ is 0, return 0 catenated to this dfn called on tail ⍵
↑,0 1∘.,⊂⍤1∇1↓⍵ ⍝ Finds 1 cat recursive call on tail ⍵ and 0 cat recursive call on ⍵.
} ⍝ Makes a matrix with a row for each possibility.
We use 1, to add a column of 1s at the start and end up with a matrix of valid partitions for ⍵.
Now the function train on the left in parens. Due to ⍨ the left argument to the train is the row of the partitions matrix and the right argument is the user input.
The train is a pile of forks with an atop as the far left tine.
((≢⊂)×1∧.=2-/10⊥¨⊂)⍨ ⍝ ⍨ swaps left and right arguments of the train.
⊂ ⍝ Partition ⍵ according to ⍺.
10⊥¨ ⍝ Decode each partition (turns strings of digits into numbers)
2-/ ⍝ Difference between adjacent cells
1∧.= ⍝ All equal 1?
⊂ ⍝ Partition ⍵ according to ⍺ again
≢ ⍝ Number of cells (ie number of partitions)
× ⍝ Multiply.
If the partition creates a sequence of descending numbers, the train returns the length of the sequence. Otherwise zero.
⍤1⊢ applies the function train between the user input and each row of the partitions matrix, returning a value for each row of the matrix. ⊢ is necessary to disambiguate between operand to ⍤ and argument to the derived function of ⍤.
⌈/ finds the maximum.
Could find a shorter algorithm but I wanted to try this way which is the most direct and declarative I could think of.
answered Jan 12 at 21:39
akhmornakhmorn
312
$endgroup$
Dyalog APL, 138 bytes
Bit of a mouthful, but it works fast for big numbers too. If you try it online, prefix the dfn by ⎕← and provide input on the right as a list of digits.
{⌈/⍵((≢⊂)×1∧.=2-/10⊥¨⊂)⍨⍤1⊢1,{⍬≡1↓⍵:↑⍬1⋄0=⊃⍵:0,∇1↓⍵⋄↑,0 1∘.,⊂⍤1∇1↓⍵}1↓⍵}
Explanation
First, the inner dfn on the right which recursively constructs a list of possible ways to partition (with ⊂) the list of digits. For example 1 0 1 0 ⊂ 2 0 1 9 returns the nested vector (2 0)(1 9).
{
⍬≡1↓⍵: ↑⍬1 ⍝ Edge case: If ⍵ is singleton list, return the column matrix (0 1)
0=⊃⍵: 0,∇1↓⍵ ⍝ If head of ⍵ is 0, return 0 catenated to this dfn called on tail ⍵
↑,0 1∘.,⊂⍤1∇1↓⍵ ⍝ Finds 1 cat recursive call on tail ⍵ and 0 cat recursive call on ⍵.
} ⍝ Makes a matrix with a row for each possibility.
We use 1, to add a column of 1s at the start and end up with a matrix of valid partitions for ⍵.
Now the function train on the left in parens. Due to ⍨ the left argument to the train is the row of the partitions matrix and the right argument is the user input.
The train is a pile of forks with an atop as the far left tine.
((≢⊂)×1∧.=2-/10⊥¨⊂)⍨ ⍝ ⍨ swaps left and right arguments of the train.
⊂ ⍝ Partition ⍵ according to ⍺.
10⊥¨ ⍝ Decode each partition (turns strings of digits into numbers)
2-/ ⍝ Difference between adjacent cells
1∧.= ⍝ All equal 1?
⊂ ⍝ Partition ⍵ according to ⍺ again
≢ ⍝ Number of cells (ie number of partitions)
× ⍝ Multiply.
If the partition creates a sequence of descending numbers, the train returns the length of the sequence. Otherwise zero.
⍤1⊢ applies the function train between the user input and each row of the partitions matrix, returning a value for each row of the matrix. ⊢ is necessary to disambiguate between operand to ⍤ and argument to the derived function of ⍤.
⌈/ finds the maximum.
Could find a shorter algorithm but I wanted to try this way which is the most direct and declarative I could think of.
answered Jan 12 at 21:39
akhmornakhmorn
312
answered Jan 12 at 21:39
akhmornakhmorn
312
answered Jan 12 at 21:39
akhmornakhmorn
312
answered Jan 12 at 21:39
akhmornakhmorn
312
312
$begingroup$
Welcome to PPCG! This is an impressive first post!
$endgroup$
– Riker
Jan 12 at 21:47
add a comment |
$begingroup$
Welcome to PPCG! This is an impressive first post!
$endgroup$
– Riker
Jan 12 at 21:47
$begingroup$
Welcome to PPCG! This is an impressive first post!
$endgroup$
– Riker
Jan 12 at 21:47
$begingroup$
Welcome to PPCG! This is an impressive first post!
$endgroup$
– Riker
Jan 12 at 21:47
add a comment |
$begingroup$
TSQL, 169 bytes
Note: this can only be executed when the input can be converted to an integer.
Recursive sql used for looping.
Golfed:
DECLARE @ varchar(max) = '1211109876';
WITH C as(SELECT left(@,row_number()over(order by 1/0))+0t,@+null z,0i
FROM spt_values UNION ALL
SELECT t-1,concat(z,t),i+1FROM C WHERE i<9)SELECT
max(i)FROM C WHERE z=@
Ungolfed:
DECLARE @ varchar(max) = '1211109876';
WITH C as
(
SELECT
left(@,row_number()over(order by 1/0))+0t,
@+null z,
0i
FROM
spt_values
UNION ALL
SELECT
t-1,
concat(z,t),
i+1
FROM C
WHERE i<9
)
SELECT max(i)
FROM C
WHERE z=@
Try it out
answered Jan 11 at 21:24
t-clausen.dkt-clausen.dk
1,834314
$endgroup$
add a comment |
$begingroup$
TSQL, 169 bytes
Note: this can only be executed when the input can be converted to an integer.
Recursive sql used for looping.
Golfed:
DECLARE @ varchar(max) = '1211109876';
WITH C as(SELECT left(@,row_number()over(order by 1/0))+0t,@+null z,0i
FROM spt_values UNION ALL
SELECT t-1,concat(z,t),i+1FROM C WHERE i<9)SELECT
max(i)FROM C WHERE z=@
Ungolfed:
DECLARE @ varchar(max) = '1211109876';
WITH C as
(
SELECT
left(@,row_number()over(order by 1/0))+0t,
@+null z,
0i
FROM
spt_values
UNION ALL
SELECT
t-1,
concat(z,t),
i+1
FROM C
WHERE i<9
)
SELECT max(i)
FROM C
WHERE z=@
Try it out
answered Jan 11 at 21:24
t-clausen.dkt-clausen.dk
1,834314
$endgroup$
add a comment |
$begingroup$
TSQL, 169 bytes
Note: this can only be executed when the input can be converted to an integer.
Recursive sql used for looping.
Golfed:
DECLARE @ varchar(max) = '1211109876';
WITH C as(SELECT left(@,row_number()over(order by 1/0))+0t,@+null z,0i
FROM spt_values UNION ALL
SELECT t-1,concat(z,t),i+1FROM C WHERE i<9)SELECT
max(i)FROM C WHERE z=@
Ungolfed:
DECLARE @ varchar(max) = '1211109876';
WITH C as
(
SELECT
left(@,row_number()over(order by 1/0))+0t,
@+null z,
0i
FROM
spt_values
UNION ALL
SELECT
t-1,
concat(z,t),
i+1
FROM C
WHERE i<9
)
SELECT max(i)
FROM C
WHERE z=@
Try it out
answered Jan 11 at 21:24
t-clausen.dkt-clausen.dk
1,834314
$endgroup$
TSQL, 169 bytes
Note: this can only be executed when the input can be converted to an integer.
Recursive sql used for looping.
Golfed:
DECLARE @ varchar(max) = '1211109876';
WITH C as(SELECT left(@,row_number()over(order by 1/0))+0t,@+null z,0i
FROM spt_values UNION ALL
SELECT t-1,concat(z,t),i+1FROM C WHERE i<9)SELECT
max(i)FROM C WHERE z=@
Ungolfed:
DECLARE @ varchar(max) = '1211109876';
WITH C as
(
SELECT
left(@,row_number()over(order by 1/0))+0t,
@+null z,
0i
FROM
spt_values
UNION ALL
SELECT
t-1,
concat(z,t),
i+1
FROM C
WHERE i<9
)
SELECT max(i)
FROM C
WHERE z=@
Try it out
answered Jan 11 at 21:24
t-clausen.dkt-clausen.dk
1,834314
edited Jan 14 at 9:02
answered Jan 11 at 21:24
t-clausen.dkt-clausen.dk
1,834314
answered Jan 11 at 21:24
t-clausen.dkt-clausen.dk
1,834314
answered Jan 11 at 21:24
t-clausen.dkt-clausen.dk
1,834314
1,834314
add a comment |
add a comment |
If this is an answer to a challenge…
…Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.
…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
Explanations of your answer make it more interesting to read and are very much encouraged.…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.
More generally…
…Please make sure to answer the question and provide sufficient detail.
…Avoid asking for help, clarification or responding to other answers (use comments instead).
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$begingroup$
Migrated from sandbox : codegolf.meta.stackexchange.com/questions/2140/…
$endgroup$
– digEmAll
Jan 4 at 10:49
1
$begingroup$
Is the test case
210 -> 2,1,0wrong (same with0 -> 0)? The tasks says "sub-numbers cannot contain leading zeros", is zero a special case?$endgroup$
– BMO
Jan 4 at 17:30
2
$begingroup$
@BMO: well, here the topic is kinda phylosofical... :D to me, 0 is a number with no (useless) leading zero, so yes zero is a special case
$endgroup$
– digEmAll
Jan 4 at 17:39
2
$begingroup$
would you call these... condescending numbers? xD sorry that was like not even funny
$endgroup$
– HyperNeutrino
Jan 6 at 3:03
$begingroup$
Sorry, deleted my comment in which I asked about
212019. Seems I didn't read all the rules.$endgroup$
– cyclaminist
Jan 12 at 21:49