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Find the starting and ending index of Smallest sum contiguous subarray in C++
0
I am trying to find the starting and ending index of the Smallest sum contiguous subarray. I tried many times but I could not able to find.
Can Anyone help me out with this in C++ ?
Code for finding Smallest sum contiguous subarray :
#include <bits/stdc++.h>
using namespace std;
int main()
{
int arr = {3, -4, 2, -3, -1, 7, -5};
int n = sizeof(arr) / sizeof(arr[0]);
int min_ending_here = INT_MAX;
int min_so_far = INT_MAX;
for (int i=0; i<n; i++)
{
if (min_ending_here > 0)
min_ending_here = arr[i];
else
min_ending_here += arr[i];
min_so_far = min(min_so_far, min_ending_here);
}
cout<<"minimum sum = "<<min_so_far;
}
Output: minimum sum = -6
c++ kadanes-algorithm
asked Nov 21 '18 at 6:19
Vishal SrivastavVishal Srivastav
728
add a comment |
0
I am trying to find the starting and ending index of the Smallest sum contiguous subarray. I tried many times but I could not able to find.
Can Anyone help me out with this in C++ ?
Code for finding Smallest sum contiguous subarray :
#include <bits/stdc++.h>
using namespace std;
int main()
{
int arr = {3, -4, 2, -3, -1, 7, -5};
int n = sizeof(arr) / sizeof(arr[0]);
int min_ending_here = INT_MAX;
int min_so_far = INT_MAX;
for (int i=0; i<n; i++)
{
if (min_ending_here > 0)
min_ending_here = arr[i];
else
min_ending_here += arr[i];
min_so_far = min(min_so_far, min_ending_here);
}
cout<<"minimum sum = "<<min_so_far;
}
Output: minimum sum = -6
c++ kadanes-algorithm
asked Nov 21 '18 at 6:19
Vishal SrivastavVishal Srivastav
728
1
What makes you think-6is not the right answer ? The problem is most likely that you output the sum rather than the indexes.
– Sid S
Nov 21 '18 at 6:22
-6 is the right answer. but i want to find starting and ending index of this smallest sum contiguous subarray
– Vishal Srivastav
Nov 21 '18 at 6:24
Right, so keep track of the current starting index and the current ending index.
– Sid S
Nov 21 '18 at 6:25
1
Your code shows no attempts to get the indexes. Can you show one of your trials? Further, it would be nice to cite the source of this code :) ... geeksforgeeks.org/smallest-sum-contiguous-subarray
– Everyone
Nov 21 '18 at 6:27
add a comment |
0
0
0
I am trying to find the starting and ending index of the Smallest sum contiguous subarray. I tried many times but I could not able to find.
Can Anyone help me out with this in C++ ?
Code for finding Smallest sum contiguous subarray :
#include <bits/stdc++.h>
using namespace std;
int main()
{
int arr = {3, -4, 2, -3, -1, 7, -5};
int n = sizeof(arr) / sizeof(arr[0]);
int min_ending_here = INT_MAX;
int min_so_far = INT_MAX;
for (int i=0; i<n; i++)
{
if (min_ending_here > 0)
min_ending_here = arr[i];
else
min_ending_here += arr[i];
min_so_far = min(min_so_far, min_ending_here);
}
cout<<"minimum sum = "<<min_so_far;
}
Output: minimum sum = -6
c++ kadanes-algorithm
asked Nov 21 '18 at 6:19
Vishal SrivastavVishal Srivastav
728
I am trying to find the starting and ending index of the Smallest sum contiguous subarray. I tried many times but I could not able to find.
Can Anyone help me out with this in C++ ?
Code for finding Smallest sum contiguous subarray :
#include <bits/stdc++.h>
using namespace std;
int main()
{
int arr = {3, -4, 2, -3, -1, 7, -5};
int n = sizeof(arr) / sizeof(arr[0]);
int min_ending_here = INT_MAX;
int min_so_far = INT_MAX;
for (int i=0; i<n; i++)
{
if (min_ending_here > 0)
min_ending_here = arr[i];
else
min_ending_here += arr[i];
min_so_far = min(min_so_far, min_ending_here);
}
cout<<"minimum sum = "<<min_so_far;
}
Output: minimum sum = -6
c++ kadanes-algorithm
c++ kadanes-algorithm
asked Nov 21 '18 at 6:19
Vishal SrivastavVishal Srivastav
728
asked Nov 21 '18 at 6:19
Vishal SrivastavVishal Srivastav
728
asked Nov 21 '18 at 6:19
Vishal SrivastavVishal Srivastav
728
asked Nov 21 '18 at 6:19
Vishal SrivastavVishal Srivastav
728
asked Nov 21 '18 at 6:19
Vishal SrivastavVishal Srivastav
728
728
1
What makes you think-6is not the right answer ? The problem is most likely that you output the sum rather than the indexes.
– Sid S
Nov 21 '18 at 6:22
-6 is the right answer. but i want to find starting and ending index of this smallest sum contiguous subarray
– Vishal Srivastav
Nov 21 '18 at 6:24
Right, so keep track of the current starting index and the current ending index.
– Sid S
Nov 21 '18 at 6:25
1
Your code shows no attempts to get the indexes. Can you show one of your trials? Further, it would be nice to cite the source of this code :) ... geeksforgeeks.org/smallest-sum-contiguous-subarray
– Everyone
Nov 21 '18 at 6:27
add a comment |
1
What makes you think-6is not the right answer ? The problem is most likely that you output the sum rather than the indexes.
– Sid S
Nov 21 '18 at 6:22
-6 is the right answer. but i want to find starting and ending index of this smallest sum contiguous subarray
– Vishal Srivastav
Nov 21 '18 at 6:24
Right, so keep track of the current starting index and the current ending index.
– Sid S
Nov 21 '18 at 6:25
1
Your code shows no attempts to get the indexes. Can you show one of your trials? Further, it would be nice to cite the source of this code :) ... geeksforgeeks.org/smallest-sum-contiguous-subarray
– Everyone
Nov 21 '18 at 6:27
1
1
What makes you think
-6 is not the right answer ? The problem is most likely that you output the sum rather than the indexes.– Sid S
Nov 21 '18 at 6:22
What makes you think
-6 is not the right answer ? The problem is most likely that you output the sum rather than the indexes.– Sid S
Nov 21 '18 at 6:22
-6 is the right answer. but i want to find starting and ending index of this smallest sum contiguous subarray
– Vishal Srivastav
Nov 21 '18 at 6:24
-6 is the right answer. but i want to find starting and ending index of this smallest sum contiguous subarray
– Vishal Srivastav
Nov 21 '18 at 6:24
Right, so keep track of the current starting index and the current ending index.
– Sid S
Nov 21 '18 at 6:25
Right, so keep track of the current starting index and the current ending index.
– Sid S
Nov 21 '18 at 6:25
1
1
Your code shows no attempts to get the indexes. Can you show one of your trials? Further, it would be nice to cite the source of this code :) ... geeksforgeeks.org/smallest-sum-contiguous-subarray
– Everyone
Nov 21 '18 at 6:27
Your code shows no attempts to get the indexes. Can you show one of your trials? Further, it would be nice to cite the source of this code :) ... geeksforgeeks.org/smallest-sum-contiguous-subarray
– Everyone
Nov 21 '18 at 6:27
add a comment |
2 Answers
2
active
oldest
votes
2
Assuming that if there exists multiple contiguous sub-arrays with minimum sum we will take the one with minimum starting index, we may modify above solution as below:
#include <bits/stdc++.h>
using namespace std;
int main() {
int arr = {3, -4, 2, -3, -1, 7, -5};
int n = sizeof(arr) / sizeof(arr[0]);
int min_ending_here = INT_MAX;
int min_so_far = INT_MAX;
int last_idx = 0; // indication of fresh start of contiguous summation
int start_idx; // for holding start index
int end_idx; // for holding end index
for (int i=0; i<n; i++) {
if (min_ending_here > 0) {
min_ending_here = arr[i];
last_idx = i;
}
else {
min_ending_here += arr[i];
}
if (min_so_far > min_ending_here) {
min_so_far = min_ending_here;
start_idx = last_idx;
end_idx = i;
}
}
cout<<"minimum sum = "<<min_so_far<<endl;
cout<<"start index = "<<start_idx<<endl;
cout<<"end index = "<<end_idx<<endl;
}
answered Nov 21 '18 at 9:45
Bishal GautamBishal Gautam
810517
@Vishal Srivastav, If above solution helped you, you may mark this as Accepted so that it may help future users in similar problem. Thanks !!
– Bishal Gautam
Nov 21 '18 at 11:48
add a comment |
1
I tested it with 2 arrays (current ( the result is [1,4]) and commented (the result is [3,3]), see the code):
#include <bits/stdc++.h>
using namespace std;
int main()
{
int arr = {3, -4, 2, -3, -1, 7, -5};
//int arr = {2, 6, 8, 1, 4};
int n = sizeof(arr) / sizeof(arr[0]);
int min_ending_here = INT_MAX;
int min_so_far = INT_MAX;
int infimum = INT_MAX; // hold minimum value.
std::pair<int, int> idx(-1, -1); // subarray's indexes
for (int i=0; i<n; i++)
{
if (min_ending_here > 0)
{
min_ending_here = arr[i];
}
else
min_ending_here += arr[i];
min_so_far = min(min_so_far, min_ending_here);
// indexes addition
if( min_so_far == min_ending_here)
{
infimum = min(arr[i], infimum);
if( infimum == arr[i] )
{
idx.first = i;
}
idx.second = i;
}
// << indexes addition
}
cout<<"minimum sum = "<<min_so_far << " indexes: " << idx.first << " " << idx.second;
}
answered Nov 21 '18 at 7:15
Alexander CherninAlexander Chernin
307210
wrong answer for arr = {3, -4, 2, -3, -1, -7, -5}; your output is : minimum sum = -18 ( sum is correct ) BUT indexes: 5 6 ( indices are wrong )
– Vishal Srivastav
Nov 21 '18 at 8:19
1 upvote for your effort..because no one responded..
– Vishal Srivastav
Nov 21 '18 at 8:20
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
Assuming that if there exists multiple contiguous sub-arrays with minimum sum we will take the one with minimum starting index, we may modify above solution as below:
#include <bits/stdc++.h>
using namespace std;
int main() {
int arr = {3, -4, 2, -3, -1, 7, -5};
int n = sizeof(arr) / sizeof(arr[0]);
int min_ending_here = INT_MAX;
int min_so_far = INT_MAX;
int last_idx = 0; // indication of fresh start of contiguous summation
int start_idx; // for holding start index
int end_idx; // for holding end index
for (int i=0; i<n; i++) {
if (min_ending_here > 0) {
min_ending_here = arr[i];
last_idx = i;
}
else {
min_ending_here += arr[i];
}
if (min_so_far > min_ending_here) {
min_so_far = min_ending_here;
start_idx = last_idx;
end_idx = i;
}
}
cout<<"minimum sum = "<<min_so_far<<endl;
cout<<"start index = "<<start_idx<<endl;
cout<<"end index = "<<end_idx<<endl;
}
answered Nov 21 '18 at 9:45
Bishal GautamBishal Gautam
810517
@Vishal Srivastav, If above solution helped you, you may mark this as Accepted so that it may help future users in similar problem. Thanks !!
– Bishal Gautam
Nov 21 '18 at 11:48
add a comment |
2
Assuming that if there exists multiple contiguous sub-arrays with minimum sum we will take the one with minimum starting index, we may modify above solution as below:
#include <bits/stdc++.h>
using namespace std;
int main() {
int arr = {3, -4, 2, -3, -1, 7, -5};
int n = sizeof(arr) / sizeof(arr[0]);
int min_ending_here = INT_MAX;
int min_so_far = INT_MAX;
int last_idx = 0; // indication of fresh start of contiguous summation
int start_idx; // for holding start index
int end_idx; // for holding end index
for (int i=0; i<n; i++) {
if (min_ending_here > 0) {
min_ending_here = arr[i];
last_idx = i;
}
else {
min_ending_here += arr[i];
}
if (min_so_far > min_ending_here) {
min_so_far = min_ending_here;
start_idx = last_idx;
end_idx = i;
}
}
cout<<"minimum sum = "<<min_so_far<<endl;
cout<<"start index = "<<start_idx<<endl;
cout<<"end index = "<<end_idx<<endl;
}
answered Nov 21 '18 at 9:45
Bishal GautamBishal Gautam
810517
@Vishal Srivastav, If above solution helped you, you may mark this as Accepted so that it may help future users in similar problem. Thanks !!
– Bishal Gautam
Nov 21 '18 at 11:48
add a comment |
2
2
2
Assuming that if there exists multiple contiguous sub-arrays with minimum sum we will take the one with minimum starting index, we may modify above solution as below:
#include <bits/stdc++.h>
using namespace std;
int main() {
int arr = {3, -4, 2, -3, -1, 7, -5};
int n = sizeof(arr) / sizeof(arr[0]);
int min_ending_here = INT_MAX;
int min_so_far = INT_MAX;
int last_idx = 0; // indication of fresh start of contiguous summation
int start_idx; // for holding start index
int end_idx; // for holding end index
for (int i=0; i<n; i++) {
if (min_ending_here > 0) {
min_ending_here = arr[i];
last_idx = i;
}
else {
min_ending_here += arr[i];
}
if (min_so_far > min_ending_here) {
min_so_far = min_ending_here;
start_idx = last_idx;
end_idx = i;
}
}
cout<<"minimum sum = "<<min_so_far<<endl;
cout<<"start index = "<<start_idx<<endl;
cout<<"end index = "<<end_idx<<endl;
}
answered Nov 21 '18 at 9:45
Bishal GautamBishal Gautam
810517
Assuming that if there exists multiple contiguous sub-arrays with minimum sum we will take the one with minimum starting index, we may modify above solution as below:
#include <bits/stdc++.h>
using namespace std;
int main() {
int arr = {3, -4, 2, -3, -1, 7, -5};
int n = sizeof(arr) / sizeof(arr[0]);
int min_ending_here = INT_MAX;
int min_so_far = INT_MAX;
int last_idx = 0; // indication of fresh start of contiguous summation
int start_idx; // for holding start index
int end_idx; // for holding end index
for (int i=0; i<n; i++) {
if (min_ending_here > 0) {
min_ending_here = arr[i];
last_idx = i;
}
else {
min_ending_here += arr[i];
}
if (min_so_far > min_ending_here) {
min_so_far = min_ending_here;
start_idx = last_idx;
end_idx = i;
}
}
cout<<"minimum sum = "<<min_so_far<<endl;
cout<<"start index = "<<start_idx<<endl;
cout<<"end index = "<<end_idx<<endl;
}
answered Nov 21 '18 at 9:45
Bishal GautamBishal Gautam
810517
answered Nov 21 '18 at 9:45
Bishal GautamBishal Gautam
810517
answered Nov 21 '18 at 9:45
Bishal GautamBishal Gautam
810517
answered Nov 21 '18 at 9:45
Bishal GautamBishal Gautam
810517
810517
@Vishal Srivastav, If above solution helped you, you may mark this as Accepted so that it may help future users in similar problem. Thanks !!
– Bishal Gautam
Nov 21 '18 at 11:48
add a comment |
@Vishal Srivastav, If above solution helped you, you may mark this as Accepted so that it may help future users in similar problem. Thanks !!
– Bishal Gautam
Nov 21 '18 at 11:48
@Vishal Srivastav, If above solution helped you, you may mark this as Accepted so that it may help future users in similar problem. Thanks !!
– Bishal Gautam
Nov 21 '18 at 11:48
@Vishal Srivastav, If above solution helped you, you may mark this as Accepted so that it may help future users in similar problem. Thanks !!
– Bishal Gautam
Nov 21 '18 at 11:48
add a comment |
1
I tested it with 2 arrays (current ( the result is [1,4]) and commented (the result is [3,3]), see the code):
#include <bits/stdc++.h>
using namespace std;
int main()
{
int arr = {3, -4, 2, -3, -1, 7, -5};
//int arr = {2, 6, 8, 1, 4};
int n = sizeof(arr) / sizeof(arr[0]);
int min_ending_here = INT_MAX;
int min_so_far = INT_MAX;
int infimum = INT_MAX; // hold minimum value.
std::pair<int, int> idx(-1, -1); // subarray's indexes
for (int i=0; i<n; i++)
{
if (min_ending_here > 0)
{
min_ending_here = arr[i];
}
else
min_ending_here += arr[i];
min_so_far = min(min_so_far, min_ending_here);
// indexes addition
if( min_so_far == min_ending_here)
{
infimum = min(arr[i], infimum);
if( infimum == arr[i] )
{
idx.first = i;
}
idx.second = i;
}
// << indexes addition
}
cout<<"minimum sum = "<<min_so_far << " indexes: " << idx.first << " " << idx.second;
}
answered Nov 21 '18 at 7:15
Alexander CherninAlexander Chernin
307210
wrong answer for arr = {3, -4, 2, -3, -1, -7, -5}; your output is : minimum sum = -18 ( sum is correct ) BUT indexes: 5 6 ( indices are wrong )
– Vishal Srivastav
Nov 21 '18 at 8:19
1 upvote for your effort..because no one responded..
– Vishal Srivastav
Nov 21 '18 at 8:20
add a comment |
1
I tested it with 2 arrays (current ( the result is [1,4]) and commented (the result is [3,3]), see the code):
#include <bits/stdc++.h>
using namespace std;
int main()
{
int arr = {3, -4, 2, -3, -1, 7, -5};
//int arr = {2, 6, 8, 1, 4};
int n = sizeof(arr) / sizeof(arr[0]);
int min_ending_here = INT_MAX;
int min_so_far = INT_MAX;
int infimum = INT_MAX; // hold minimum value.
std::pair<int, int> idx(-1, -1); // subarray's indexes
for (int i=0; i<n; i++)
{
if (min_ending_here > 0)
{
min_ending_here = arr[i];
}
else
min_ending_here += arr[i];
min_so_far = min(min_so_far, min_ending_here);
// indexes addition
if( min_so_far == min_ending_here)
{
infimum = min(arr[i], infimum);
if( infimum == arr[i] )
{
idx.first = i;
}
idx.second = i;
}
// << indexes addition
}
cout<<"minimum sum = "<<min_so_far << " indexes: " << idx.first << " " << idx.second;
}
answered Nov 21 '18 at 7:15
Alexander CherninAlexander Chernin
307210
wrong answer for arr = {3, -4, 2, -3, -1, -7, -5}; your output is : minimum sum = -18 ( sum is correct ) BUT indexes: 5 6 ( indices are wrong )
– Vishal Srivastav
Nov 21 '18 at 8:19
1 upvote for your effort..because no one responded..
– Vishal Srivastav
Nov 21 '18 at 8:20
add a comment |
1
1
1
I tested it with 2 arrays (current ( the result is [1,4]) and commented (the result is [3,3]), see the code):
#include <bits/stdc++.h>
using namespace std;
int main()
{
int arr = {3, -4, 2, -3, -1, 7, -5};
//int arr = {2, 6, 8, 1, 4};
int n = sizeof(arr) / sizeof(arr[0]);
int min_ending_here = INT_MAX;
int min_so_far = INT_MAX;
int infimum = INT_MAX; // hold minimum value.
std::pair<int, int> idx(-1, -1); // subarray's indexes
for (int i=0; i<n; i++)
{
if (min_ending_here > 0)
{
min_ending_here = arr[i];
}
else
min_ending_here += arr[i];
min_so_far = min(min_so_far, min_ending_here);
// indexes addition
if( min_so_far == min_ending_here)
{
infimum = min(arr[i], infimum);
if( infimum == arr[i] )
{
idx.first = i;
}
idx.second = i;
}
// << indexes addition
}
cout<<"minimum sum = "<<min_so_far << " indexes: " << idx.first << " " << idx.second;
}
answered Nov 21 '18 at 7:15
Alexander CherninAlexander Chernin
307210
I tested it with 2 arrays (current ( the result is [1,4]) and commented (the result is [3,3]), see the code):
#include <bits/stdc++.h>
using namespace std;
int main()
{
int arr = {3, -4, 2, -3, -1, 7, -5};
//int arr = {2, 6, 8, 1, 4};
int n = sizeof(arr) / sizeof(arr[0]);
int min_ending_here = INT_MAX;
int min_so_far = INT_MAX;
int infimum = INT_MAX; // hold minimum value.
std::pair<int, int> idx(-1, -1); // subarray's indexes
for (int i=0; i<n; i++)
{
if (min_ending_here > 0)
{
min_ending_here = arr[i];
}
else
min_ending_here += arr[i];
min_so_far = min(min_so_far, min_ending_here);
// indexes addition
if( min_so_far == min_ending_here)
{
infimum = min(arr[i], infimum);
if( infimum == arr[i] )
{
idx.first = i;
}
idx.second = i;
}
// << indexes addition
}
cout<<"minimum sum = "<<min_so_far << " indexes: " << idx.first << " " << idx.second;
}
answered Nov 21 '18 at 7:15
Alexander CherninAlexander Chernin
307210
answered Nov 21 '18 at 7:15
Alexander CherninAlexander Chernin
307210
answered Nov 21 '18 at 7:15
Alexander CherninAlexander Chernin
307210
answered Nov 21 '18 at 7:15
Alexander CherninAlexander Chernin
307210
307210
wrong answer for arr = {3, -4, 2, -3, -1, -7, -5}; your output is : minimum sum = -18 ( sum is correct ) BUT indexes: 5 6 ( indices are wrong )
– Vishal Srivastav
Nov 21 '18 at 8:19
1 upvote for your effort..because no one responded..
– Vishal Srivastav
Nov 21 '18 at 8:20
add a comment |
wrong answer for arr = {3, -4, 2, -3, -1, -7, -5}; your output is : minimum sum = -18 ( sum is correct ) BUT indexes: 5 6 ( indices are wrong )
– Vishal Srivastav
Nov 21 '18 at 8:19
1 upvote for your effort..because no one responded..
– Vishal Srivastav
Nov 21 '18 at 8:20
wrong answer for arr = {3, -4, 2, -3, -1, -7, -5}; your output is : minimum sum = -18 ( sum is correct ) BUT indexes: 5 6 ( indices are wrong )
– Vishal Srivastav
Nov 21 '18 at 8:19
wrong answer for arr = {3, -4, 2, -3, -1, -7, -5}; your output is : minimum sum = -18 ( sum is correct ) BUT indexes: 5 6 ( indices are wrong )
– Vishal Srivastav
Nov 21 '18 at 8:19
1 upvote for your effort..because no one responded..
– Vishal Srivastav
Nov 21 '18 at 8:20
1 upvote for your effort..because no one responded..
– Vishal Srivastav
Nov 21 '18 at 8:20
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What makes you think
-6is not the right answer ? The problem is most likely that you output the sum rather than the indexes.– Sid S
Nov 21 '18 at 6:22
-6 is the right answer. but i want to find starting and ending index of this smallest sum contiguous subarray
– Vishal Srivastav
Nov 21 '18 at 6:24
Right, so keep track of the current starting index and the current ending index.
– Sid S
Nov 21 '18 at 6:25
1
Your code shows no attempts to get the indexes. Can you show one of your trials? Further, it would be nice to cite the source of this code :) ... geeksforgeeks.org/smallest-sum-contiguous-subarray
– Everyone
Nov 21 '18 at 6:27