Mixing
Get the Strings that occur exactly three times from Arraylist
I have an ArrayList which contains some values with duplicates and elements that occur thrice, I want to collect those values that occur thrice specifically into another ArrayList like
Arraylist<String> strings; //contains all strings that are duplicates and that occur thrice
Here, I want to get only the Strings that occur thrice in another array list.
Arraylist<String> thrice; //contains only elements that occur three times.
Currently, I have a solution for dealing with duplicates but I cannot extend this for only getting strings that occur thrice, this please help me to find out.
java list arraylist
edited Jan 4 at 8:34
Naman
45.1k11102204
asked Jan 2 at 16:29
Vipul ChauhanVipul Chauhan
383
add a comment |
I have an ArrayList which contains some values with duplicates and elements that occur thrice, I want to collect those values that occur thrice specifically into another ArrayList like
Arraylist<String> strings; //contains all strings that are duplicates and that occur thrice
Here, I want to get only the Strings that occur thrice in another array list.
Arraylist<String> thrice; //contains only elements that occur three times.
Currently, I have a solution for dealing with duplicates but I cannot extend this for only getting strings that occur thrice, this please help me to find out.
java list arraylist
edited Jan 4 at 8:34
Naman
45.1k11102204
asked Jan 2 at 16:29
Vipul ChauhanVipul Chauhan
383
1
Basically, instead of using a HashSet you now want to use a HashMap to count the number of occurrences of each string.
– shash678
Jan 2 at 16:31
See this answer for a solution as @shash678 suggested
– Stalemate Of Tuning
Jan 2 at 16:35
Possible duplicate of get the duplicates values from Arraylist<String> and then get those items in another Arraylist
– Vipul Chauhan
Jan 4 at 17:25
add a comment |
I have an ArrayList which contains some values with duplicates and elements that occur thrice, I want to collect those values that occur thrice specifically into another ArrayList like
Arraylist<String> strings; //contains all strings that are duplicates and that occur thrice
Here, I want to get only the Strings that occur thrice in another array list.
Arraylist<String> thrice; //contains only elements that occur three times.
Currently, I have a solution for dealing with duplicates but I cannot extend this for only getting strings that occur thrice, this please help me to find out.
java list arraylist
edited Jan 4 at 8:34
Naman
45.1k11102204
asked Jan 2 at 16:29
Vipul ChauhanVipul Chauhan
383
I have an ArrayList which contains some values with duplicates and elements that occur thrice, I want to collect those values that occur thrice specifically into another ArrayList like
Arraylist<String> strings; //contains all strings that are duplicates and that occur thrice
Here, I want to get only the Strings that occur thrice in another array list.
Arraylist<String> thrice; //contains only elements that occur three times.
Currently, I have a solution for dealing with duplicates but I cannot extend this for only getting strings that occur thrice, this please help me to find out.
java list arraylist
java list arraylist
edited Jan 4 at 8:34
Naman
45.1k11102204
asked Jan 2 at 16:29
Vipul ChauhanVipul Chauhan
383
edited Jan 4 at 8:34
Naman
45.1k11102204
asked Jan 2 at 16:29
Vipul ChauhanVipul Chauhan
383
edited Jan 4 at 8:34
Naman
45.1k11102204
edited Jan 4 at 8:34
Naman
45.1k11102204
edited Jan 4 at 8:34
Naman
45.1k11102204
45.1k11102204
asked Jan 2 at 16:29
Vipul ChauhanVipul Chauhan
383
asked Jan 2 at 16:29
Vipul ChauhanVipul Chauhan
383
asked Jan 2 at 16:29
Vipul ChauhanVipul Chauhan
383
383
1
Basically, instead of using a HashSet you now want to use a HashMap to count the number of occurrences of each string.
– shash678
Jan 2 at 16:31
See this answer for a solution as @shash678 suggested
– Stalemate Of Tuning
Jan 2 at 16:35
Possible duplicate of get the duplicates values from Arraylist<String> and then get those items in another Arraylist
– Vipul Chauhan
Jan 4 at 17:25
add a comment |
1
Basically, instead of using a HashSet you now want to use a HashMap to count the number of occurrences of each string.
– shash678
Jan 2 at 16:31
See this answer for a solution as @shash678 suggested
– Stalemate Of Tuning
Jan 2 at 16:35
Possible duplicate of get the duplicates values from Arraylist<String> and then get those items in another Arraylist
– Vipul Chauhan
Jan 4 at 17:25
1
1
Basically, instead of using a HashSet you now want to use a HashMap to count the number of occurrences of each string.
– shash678
Jan 2 at 16:31
Basically, instead of using a HashSet you now want to use a HashMap to count the number of occurrences of each string.
– shash678
Jan 2 at 16:31
See this answer for a solution as @shash678 suggested
– Stalemate Of Tuning
Jan 2 at 16:35
See this answer for a solution as @shash678 suggested
– Stalemate Of Tuning
Jan 2 at 16:35
Possible duplicate of get the duplicates values from Arraylist<String> and then get those items in another Arraylist
– Vipul Chauhan
Jan 4 at 17:25
Possible duplicate of get the duplicates values from Arraylist<String> and then get those items in another Arraylist
– Vipul Chauhan
Jan 4 at 17:25
add a comment |
5 Answers
5
active
oldest
votes
A generic utility of what you're trying to acheieve in both the questions would be using Collections.frequency as :
/**
* @param input the list as your input
* @param n number of occurrence (duplicates:2 , triplets:3 etc..)
* @param <T> (type of elements)
* @return elements with such conditional occurrent in a Set
*/
static <T> Set<T> findElementsWithNOccurrence(List<T> input, int n) {
return input.stream()
.filter(a -> Collections.frequency(input, a) == n) // filter by number of occurrences
.collect(Collectors.toSet()); // collecting to a final set (one representation of each)
}
Note: This would be an O(n^2) approach since its using Collections.frequency which iterates over the entire collection again to get the frequency. But proposed for a more readable and generic approach towards what you're looking for. Also, this intentionally collects final output to a Set, since a List can again have duplicates after all.
Alternatively, you can use the method to count the frequency of elements in Java-8 and iterate over the entries of the Map created thereby to process filtering as desired and collect the output in the same iteration :
/**
* @param input the list as your input
* @param n number of occurrence (duplicates :2 , triplets :3 etc)
* @param <T> (type of elements)
* @return elements in a set
*/
static <T> Set<T> findElementsWithNOccurrence(List<T> input, int n) {
return input.stream() // Stream<T>
.collect(Collectors.groupingBy(Function.identity(),
Collectors.counting())) // Map<T, Long>
.entrySet() // Set<Map.Entry<T,Long>>
.stream() // Stream<Map.Entry<T,Long>>
.filter(e -> e.getValue() == n) // filtered with frequency 'n'
.map(Map.Entry::getKey) // Stream<T>
.collect(Collectors.toSet()); // collect to Set
}
answered Jan 2 at 17:10
NamanNaman
45.1k11102204
add a comment |
You can do it via a stream as follows:
List<String> result = strings.stream()
.collect(Collectors.groupingBy(Function.identity(), counting()))
.entrySet().stream()
.filter(e -> e.getValue() == 3) // keep only elements that occur 3 times
.map(Map.Entry::getKey)
.collect(Collectors.toList());
You could also do it as follows, but I'd recommend the above as it's more preferable.
List<String> result = new HashSet<>(strings).stream()
.filter(item -> strings.stream()
.filter(e -> e.equals(item)).limit(3).count() == 3)
.collect(Collectors.toList());
answered Jan 2 at 16:36
AomineAomine
42.6k74677
1
The second one doesn't seem really efficient, it has a O(n^2) time complexity for a problem which is easily solved in O(n), kind of overkill.
– Ricola
Jan 2 at 16:47
@Ricola First is definitely preferable. I'll edit to explicitly say so.
– Aomine
Jan 2 at 16:49
add a comment |
Basically, instead of using a HashSet you now want to use a HashMap to count the number of occurrences of each string.
Furthermore, instead of writing a method for finding the strings that occur three times specifically, you could write a method that takes in a parameter, n and finds the strings that occur N times:
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.Set;
class StackOverflowQ {
static List<String> getStringsThatOccurNTimes(List<String> stringList, int n) {
if (stringList == null || stringList.size() == 0) {
return stringList;
}
Set<String> stringsThatOccurNTimesSet = new HashSet<>();
Map<String, Integer> stringCounts = new HashMap<>();
for (String s : stringList) {
int currentStringCount = stringCounts.getOrDefault(s, 0) + 1;
stringCounts.put(s, currentStringCount);
if (currentStringCount == n) {
stringsThatOccurNTimesSet.add(s);
} else if (currentStringCount == n + 1) {
stringsThatOccurNTimesSet.remove(s); // We use a set so this operation is O(1)
}
}
return new ArrayList<>(stringsThatOccurNTimesSet);
}
public static void main(String args) {
List<String> stringsList = new ArrayList<>(Arrays.asList("a", "b", "c", "d", "e", "b", "c", "c", "d", "d", "d", "e"));
List<String> stringsThatOccurTwoTimes = getStringsThatOccurNTimes(stringsList, 2);
List<String> stringsThatOccurThreeTimes = getStringsThatOccurNTimes(stringsList, 3);
List<String> stringsThatOccurFourTimes = getStringsThatOccurNTimes(stringsList, 4);
System.out.println("Original list: " + stringsList);
System.out.println("Strings that occur two times: " + stringsThatOccurTwoTimes);
System.out.println("Strings that occur three times: " + stringsThatOccurThreeTimes);
System.out.println("Strings that occur four times: " + stringsThatOccurFourTimes);
}
}
Output:
Original list: [a, b, c, d, e, b, c, c, d, d, d, e]
Strings that occur two times: [b, e]
Strings that occur three times: [c]
Strings that occur four times: [d]
answered Jan 2 at 16:35
shash678shash678
5,93521030
How about a util for this in general?
– Naman
Jan 2 at 17:13
Nice I use something similar in python when I do interview questions that are similar to OP's one.
– shash678
Jan 2 at 17:28
add a comment |
The best way to do this is.....
Making of required arraylist.........
ArrayList<String> thrice=new ArrayList<>();
ArrayList<String> one=new ArrayList<>();
Adding some values for checking.......
one.add("1");one.add("1");one.add("1");one.add("1");one.add("1");
one.add("2");one.add("2");one.add("2");one.add("2");one.add("2");
one.add("1");one.add("1");one.add("1");
one.add("3");one.add("3");
one.add("4");one.add("5");
one.add("2");one.add("2");one.add("2");
Mapping to done the task.....
Map<String, Integer> duplicates = new HashMap<String, Integer>();
for (String str : one) {
if (duplicates.containsKey(str)) {
duplicates.put(str, duplicates.get(str) + 1);
} else {
duplicates.put(str, 1);
}
}
for (Map.Entry<String, Integer> entry : duplicates.entrySet()) {
if(entry.getValue()>=3){
thrice.add(entry.getKey());
}
}
Set the list in listview...
ArrayAdapter<String> ad=new ArrayAdapter<>
(this,android.R.layout.simple_list_item_1,thrice);
ListView lv=findViewById(R.id.new_l);
lv.setAdapter(ad);
Output= 1,2
Already check the answer and you can change the condition according to your way
if(entry.getValue()>=3){
thrice.add(entry.getKey());
}
Easiest answer from above...not required to change the versions of minimum sdk for android user
answered Jan 5 at 7:10
Vipul ChauhanVipul Chauhan
4310
add a comment |
You can use computeIfAbsent for this.
List<String> strings;
final Map<String, BigInteger> stringCounts = new HashMap<>();
strings.stream().forEach(s -> {
BigInteger count = stringCounts.computeIfAbsent(s, k -> BigInteger.ZERO).add(BigInteger.ONE);
stringCounts.put(s, count);
});
Now filter stringCounts with value=3.
answered Jan 2 at 17:06
fastcodejavafastcodejava
24.6k19109162
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
A generic utility of what you're trying to acheieve in both the questions would be using Collections.frequency as :
/**
* @param input the list as your input
* @param n number of occurrence (duplicates:2 , triplets:3 etc..)
* @param <T> (type of elements)
* @return elements with such conditional occurrent in a Set
*/
static <T> Set<T> findElementsWithNOccurrence(List<T> input, int n) {
return input.stream()
.filter(a -> Collections.frequency(input, a) == n) // filter by number of occurrences
.collect(Collectors.toSet()); // collecting to a final set (one representation of each)
}
Note: This would be an O(n^2) approach since its using Collections.frequency which iterates over the entire collection again to get the frequency. But proposed for a more readable and generic approach towards what you're looking for. Also, this intentionally collects final output to a Set, since a List can again have duplicates after all.
Alternatively, you can use the method to count the frequency of elements in Java-8 and iterate over the entries of the Map created thereby to process filtering as desired and collect the output in the same iteration :
/**
* @param input the list as your input
* @param n number of occurrence (duplicates :2 , triplets :3 etc)
* @param <T> (type of elements)
* @return elements in a set
*/
static <T> Set<T> findElementsWithNOccurrence(List<T> input, int n) {
return input.stream() // Stream<T>
.collect(Collectors.groupingBy(Function.identity(),
Collectors.counting())) // Map<T, Long>
.entrySet() // Set<Map.Entry<T,Long>>
.stream() // Stream<Map.Entry<T,Long>>
.filter(e -> e.getValue() == n) // filtered with frequency 'n'
.map(Map.Entry::getKey) // Stream<T>
.collect(Collectors.toSet()); // collect to Set
}
answered Jan 2 at 17:10
NamanNaman
45.1k11102204
add a comment |
A generic utility of what you're trying to acheieve in both the questions would be using Collections.frequency as :
/**
* @param input the list as your input
* @param n number of occurrence (duplicates:2 , triplets:3 etc..)
* @param <T> (type of elements)
* @return elements with such conditional occurrent in a Set
*/
static <T> Set<T> findElementsWithNOccurrence(List<T> input, int n) {
return input.stream()
.filter(a -> Collections.frequency(input, a) == n) // filter by number of occurrences
.collect(Collectors.toSet()); // collecting to a final set (one representation of each)
}
Note: This would be an O(n^2) approach since its using Collections.frequency which iterates over the entire collection again to get the frequency. But proposed for a more readable and generic approach towards what you're looking for. Also, this intentionally collects final output to a Set, since a List can again have duplicates after all.
Alternatively, you can use the method to count the frequency of elements in Java-8 and iterate over the entries of the Map created thereby to process filtering as desired and collect the output in the same iteration :
/**
* @param input the list as your input
* @param n number of occurrence (duplicates :2 , triplets :3 etc)
* @param <T> (type of elements)
* @return elements in a set
*/
static <T> Set<T> findElementsWithNOccurrence(List<T> input, int n) {
return input.stream() // Stream<T>
.collect(Collectors.groupingBy(Function.identity(),
Collectors.counting())) // Map<T, Long>
.entrySet() // Set<Map.Entry<T,Long>>
.stream() // Stream<Map.Entry<T,Long>>
.filter(e -> e.getValue() == n) // filtered with frequency 'n'
.map(Map.Entry::getKey) // Stream<T>
.collect(Collectors.toSet()); // collect to Set
}
answered Jan 2 at 17:10
NamanNaman
45.1k11102204
add a comment |
A generic utility of what you're trying to acheieve in both the questions would be using Collections.frequency as :
/**
* @param input the list as your input
* @param n number of occurrence (duplicates:2 , triplets:3 etc..)
* @param <T> (type of elements)
* @return elements with such conditional occurrent in a Set
*/
static <T> Set<T> findElementsWithNOccurrence(List<T> input, int n) {
return input.stream()
.filter(a -> Collections.frequency(input, a) == n) // filter by number of occurrences
.collect(Collectors.toSet()); // collecting to a final set (one representation of each)
}
Note: This would be an O(n^2) approach since its using Collections.frequency which iterates over the entire collection again to get the frequency. But proposed for a more readable and generic approach towards what you're looking for. Also, this intentionally collects final output to a Set, since a List can again have duplicates after all.
Alternatively, you can use the method to count the frequency of elements in Java-8 and iterate over the entries of the Map created thereby to process filtering as desired and collect the output in the same iteration :
/**
* @param input the list as your input
* @param n number of occurrence (duplicates :2 , triplets :3 etc)
* @param <T> (type of elements)
* @return elements in a set
*/
static <T> Set<T> findElementsWithNOccurrence(List<T> input, int n) {
return input.stream() // Stream<T>
.collect(Collectors.groupingBy(Function.identity(),
Collectors.counting())) // Map<T, Long>
.entrySet() // Set<Map.Entry<T,Long>>
.stream() // Stream<Map.Entry<T,Long>>
.filter(e -> e.getValue() == n) // filtered with frequency 'n'
.map(Map.Entry::getKey) // Stream<T>
.collect(Collectors.toSet()); // collect to Set
}
answered Jan 2 at 17:10
NamanNaman
45.1k11102204
A generic utility of what you're trying to acheieve in both the questions would be using Collections.frequency as :
/**
* @param input the list as your input
* @param n number of occurrence (duplicates:2 , triplets:3 etc..)
* @param <T> (type of elements)
* @return elements with such conditional occurrent in a Set
*/
static <T> Set<T> findElementsWithNOccurrence(List<T> input, int n) {
return input.stream()
.filter(a -> Collections.frequency(input, a) == n) // filter by number of occurrences
.collect(Collectors.toSet()); // collecting to a final set (one representation of each)
}
Note: This would be an O(n^2) approach since its using Collections.frequency which iterates over the entire collection again to get the frequency. But proposed for a more readable and generic approach towards what you're looking for. Also, this intentionally collects final output to a Set, since a List can again have duplicates after all.
Alternatively, you can use the method to count the frequency of elements in Java-8 and iterate over the entries of the Map created thereby to process filtering as desired and collect the output in the same iteration :
/**
* @param input the list as your input
* @param n number of occurrence (duplicates :2 , triplets :3 etc)
* @param <T> (type of elements)
* @return elements in a set
*/
static <T> Set<T> findElementsWithNOccurrence(List<T> input, int n) {
return input.stream() // Stream<T>
.collect(Collectors.groupingBy(Function.identity(),
Collectors.counting())) // Map<T, Long>
.entrySet() // Set<Map.Entry<T,Long>>
.stream() // Stream<Map.Entry<T,Long>>
.filter(e -> e.getValue() == n) // filtered with frequency 'n'
.map(Map.Entry::getKey) // Stream<T>
.collect(Collectors.toSet()); // collect to Set
}
answered Jan 2 at 17:10
NamanNaman
45.1k11102204
edited Jan 2 at 17:37
answered Jan 2 at 17:10
NamanNaman
45.1k11102204
answered Jan 2 at 17:10
NamanNaman
45.1k11102204
answered Jan 2 at 17:10
NamanNaman
45.1k11102204
45.1k11102204
add a comment |
add a comment |
You can do it via a stream as follows:
List<String> result = strings.stream()
.collect(Collectors.groupingBy(Function.identity(), counting()))
.entrySet().stream()
.filter(e -> e.getValue() == 3) // keep only elements that occur 3 times
.map(Map.Entry::getKey)
.collect(Collectors.toList());
You could also do it as follows, but I'd recommend the above as it's more preferable.
List<String> result = new HashSet<>(strings).stream()
.filter(item -> strings.stream()
.filter(e -> e.equals(item)).limit(3).count() == 3)
.collect(Collectors.toList());
answered Jan 2 at 16:36
AomineAomine
42.6k74677
1
The second one doesn't seem really efficient, it has a O(n^2) time complexity for a problem which is easily solved in O(n), kind of overkill.
– Ricola
Jan 2 at 16:47
@Ricola First is definitely preferable. I'll edit to explicitly say so.
– Aomine
Jan 2 at 16:49
add a comment |
You can do it via a stream as follows:
List<String> result = strings.stream()
.collect(Collectors.groupingBy(Function.identity(), counting()))
.entrySet().stream()
.filter(e -> e.getValue() == 3) // keep only elements that occur 3 times
.map(Map.Entry::getKey)
.collect(Collectors.toList());
You could also do it as follows, but I'd recommend the above as it's more preferable.
List<String> result = new HashSet<>(strings).stream()
.filter(item -> strings.stream()
.filter(e -> e.equals(item)).limit(3).count() == 3)
.collect(Collectors.toList());
answered Jan 2 at 16:36
AomineAomine
42.6k74677
1
The second one doesn't seem really efficient, it has a O(n^2) time complexity for a problem which is easily solved in O(n), kind of overkill.
– Ricola
Jan 2 at 16:47
@Ricola First is definitely preferable. I'll edit to explicitly say so.
– Aomine
Jan 2 at 16:49
add a comment |
You can do it via a stream as follows:
List<String> result = strings.stream()
.collect(Collectors.groupingBy(Function.identity(), counting()))
.entrySet().stream()
.filter(e -> e.getValue() == 3) // keep only elements that occur 3 times
.map(Map.Entry::getKey)
.collect(Collectors.toList());
You could also do it as follows, but I'd recommend the above as it's more preferable.
List<String> result = new HashSet<>(strings).stream()
.filter(item -> strings.stream()
.filter(e -> e.equals(item)).limit(3).count() == 3)
.collect(Collectors.toList());
answered Jan 2 at 16:36
AomineAomine
42.6k74677
You can do it via a stream as follows:
List<String> result = strings.stream()
.collect(Collectors.groupingBy(Function.identity(), counting()))
.entrySet().stream()
.filter(e -> e.getValue() == 3) // keep only elements that occur 3 times
.map(Map.Entry::getKey)
.collect(Collectors.toList());
You could also do it as follows, but I'd recommend the above as it's more preferable.
List<String> result = new HashSet<>(strings).stream()
.filter(item -> strings.stream()
.filter(e -> e.equals(item)).limit(3).count() == 3)
.collect(Collectors.toList());
answered Jan 2 at 16:36
AomineAomine
42.6k74677
edited Jan 2 at 16:51
answered Jan 2 at 16:36
AomineAomine
42.6k74677
answered Jan 2 at 16:36
AomineAomine
42.6k74677
answered Jan 2 at 16:36
AomineAomine
42.6k74677
42.6k74677
1
The second one doesn't seem really efficient, it has a O(n^2) time complexity for a problem which is easily solved in O(n), kind of overkill.
– Ricola
Jan 2 at 16:47
@Ricola First is definitely preferable. I'll edit to explicitly say so.
– Aomine
Jan 2 at 16:49
add a comment |
1
The second one doesn't seem really efficient, it has a O(n^2) time complexity for a problem which is easily solved in O(n), kind of overkill.
– Ricola
Jan 2 at 16:47
@Ricola First is definitely preferable. I'll edit to explicitly say so.
– Aomine
Jan 2 at 16:49
1
1
The second one doesn't seem really efficient, it has a O(n^2) time complexity for a problem which is easily solved in O(n), kind of overkill.
– Ricola
Jan 2 at 16:47
The second one doesn't seem really efficient, it has a O(n^2) time complexity for a problem which is easily solved in O(n), kind of overkill.
– Ricola
Jan 2 at 16:47
@Ricola First is definitely preferable. I'll edit to explicitly say so.
– Aomine
Jan 2 at 16:49
@Ricola First is definitely preferable. I'll edit to explicitly say so.
– Aomine
Jan 2 at 16:49
add a comment |
Basically, instead of using a HashSet you now want to use a HashMap to count the number of occurrences of each string.
Furthermore, instead of writing a method for finding the strings that occur three times specifically, you could write a method that takes in a parameter, n and finds the strings that occur N times:
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.Set;
class StackOverflowQ {
static List<String> getStringsThatOccurNTimes(List<String> stringList, int n) {
if (stringList == null || stringList.size() == 0) {
return stringList;
}
Set<String> stringsThatOccurNTimesSet = new HashSet<>();
Map<String, Integer> stringCounts = new HashMap<>();
for (String s : stringList) {
int currentStringCount = stringCounts.getOrDefault(s, 0) + 1;
stringCounts.put(s, currentStringCount);
if (currentStringCount == n) {
stringsThatOccurNTimesSet.add(s);
} else if (currentStringCount == n + 1) {
stringsThatOccurNTimesSet.remove(s); // We use a set so this operation is O(1)
}
}
return new ArrayList<>(stringsThatOccurNTimesSet);
}
public static void main(String args) {
List<String> stringsList = new ArrayList<>(Arrays.asList("a", "b", "c", "d", "e", "b", "c", "c", "d", "d", "d", "e"));
List<String> stringsThatOccurTwoTimes = getStringsThatOccurNTimes(stringsList, 2);
List<String> stringsThatOccurThreeTimes = getStringsThatOccurNTimes(stringsList, 3);
List<String> stringsThatOccurFourTimes = getStringsThatOccurNTimes(stringsList, 4);
System.out.println("Original list: " + stringsList);
System.out.println("Strings that occur two times: " + stringsThatOccurTwoTimes);
System.out.println("Strings that occur three times: " + stringsThatOccurThreeTimes);
System.out.println("Strings that occur four times: " + stringsThatOccurFourTimes);
}
}
Output:
Original list: [a, b, c, d, e, b, c, c, d, d, d, e]
Strings that occur two times: [b, e]
Strings that occur three times: [c]
Strings that occur four times: [d]
answered Jan 2 at 16:35
shash678shash678
5,93521030
How about a util for this in general?
– Naman
Jan 2 at 17:13
Nice I use something similar in python when I do interview questions that are similar to OP's one.
– shash678
Jan 2 at 17:28
add a comment |
Basically, instead of using a HashSet you now want to use a HashMap to count the number of occurrences of each string.
Furthermore, instead of writing a method for finding the strings that occur three times specifically, you could write a method that takes in a parameter, n and finds the strings that occur N times:
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.Set;
class StackOverflowQ {
static List<String> getStringsThatOccurNTimes(List<String> stringList, int n) {
if (stringList == null || stringList.size() == 0) {
return stringList;
}
Set<String> stringsThatOccurNTimesSet = new HashSet<>();
Map<String, Integer> stringCounts = new HashMap<>();
for (String s : stringList) {
int currentStringCount = stringCounts.getOrDefault(s, 0) + 1;
stringCounts.put(s, currentStringCount);
if (currentStringCount == n) {
stringsThatOccurNTimesSet.add(s);
} else if (currentStringCount == n + 1) {
stringsThatOccurNTimesSet.remove(s); // We use a set so this operation is O(1)
}
}
return new ArrayList<>(stringsThatOccurNTimesSet);
}
public static void main(String args) {
List<String> stringsList = new ArrayList<>(Arrays.asList("a", "b", "c", "d", "e", "b", "c", "c", "d", "d", "d", "e"));
List<String> stringsThatOccurTwoTimes = getStringsThatOccurNTimes(stringsList, 2);
List<String> stringsThatOccurThreeTimes = getStringsThatOccurNTimes(stringsList, 3);
List<String> stringsThatOccurFourTimes = getStringsThatOccurNTimes(stringsList, 4);
System.out.println("Original list: " + stringsList);
System.out.println("Strings that occur two times: " + stringsThatOccurTwoTimes);
System.out.println("Strings that occur three times: " + stringsThatOccurThreeTimes);
System.out.println("Strings that occur four times: " + stringsThatOccurFourTimes);
}
}
Output:
Original list: [a, b, c, d, e, b, c, c, d, d, d, e]
Strings that occur two times: [b, e]
Strings that occur three times: [c]
Strings that occur four times: [d]
answered Jan 2 at 16:35
shash678shash678
5,93521030
How about a util for this in general?
– Naman
Jan 2 at 17:13
Nice I use something similar in python when I do interview questions that are similar to OP's one.
– shash678
Jan 2 at 17:28
add a comment |
Basically, instead of using a HashSet you now want to use a HashMap to count the number of occurrences of each string.
Furthermore, instead of writing a method for finding the strings that occur three times specifically, you could write a method that takes in a parameter, n and finds the strings that occur N times:
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.Set;
class StackOverflowQ {
static List<String> getStringsThatOccurNTimes(List<String> stringList, int n) {
if (stringList == null || stringList.size() == 0) {
return stringList;
}
Set<String> stringsThatOccurNTimesSet = new HashSet<>();
Map<String, Integer> stringCounts = new HashMap<>();
for (String s : stringList) {
int currentStringCount = stringCounts.getOrDefault(s, 0) + 1;
stringCounts.put(s, currentStringCount);
if (currentStringCount == n) {
stringsThatOccurNTimesSet.add(s);
} else if (currentStringCount == n + 1) {
stringsThatOccurNTimesSet.remove(s); // We use a set so this operation is O(1)
}
}
return new ArrayList<>(stringsThatOccurNTimesSet);
}
public static void main(String args) {
List<String> stringsList = new ArrayList<>(Arrays.asList("a", "b", "c", "d", "e", "b", "c", "c", "d", "d", "d", "e"));
List<String> stringsThatOccurTwoTimes = getStringsThatOccurNTimes(stringsList, 2);
List<String> stringsThatOccurThreeTimes = getStringsThatOccurNTimes(stringsList, 3);
List<String> stringsThatOccurFourTimes = getStringsThatOccurNTimes(stringsList, 4);
System.out.println("Original list: " + stringsList);
System.out.println("Strings that occur two times: " + stringsThatOccurTwoTimes);
System.out.println("Strings that occur three times: " + stringsThatOccurThreeTimes);
System.out.println("Strings that occur four times: " + stringsThatOccurFourTimes);
}
}
Output:
Original list: [a, b, c, d, e, b, c, c, d, d, d, e]
Strings that occur two times: [b, e]
Strings that occur three times: [c]
Strings that occur four times: [d]
answered Jan 2 at 16:35
shash678shash678
5,93521030
Basically, instead of using a HashSet you now want to use a HashMap to count the number of occurrences of each string.
Furthermore, instead of writing a method for finding the strings that occur three times specifically, you could write a method that takes in a parameter, n and finds the strings that occur N times:
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.Set;
class StackOverflowQ {
static List<String> getStringsThatOccurNTimes(List<String> stringList, int n) {
if (stringList == null || stringList.size() == 0) {
return stringList;
}
Set<String> stringsThatOccurNTimesSet = new HashSet<>();
Map<String, Integer> stringCounts = new HashMap<>();
for (String s : stringList) {
int currentStringCount = stringCounts.getOrDefault(s, 0) + 1;
stringCounts.put(s, currentStringCount);
if (currentStringCount == n) {
stringsThatOccurNTimesSet.add(s);
} else if (currentStringCount == n + 1) {
stringsThatOccurNTimesSet.remove(s); // We use a set so this operation is O(1)
}
}
return new ArrayList<>(stringsThatOccurNTimesSet);
}
public static void main(String args) {
List<String> stringsList = new ArrayList<>(Arrays.asList("a", "b", "c", "d", "e", "b", "c", "c", "d", "d", "d", "e"));
List<String> stringsThatOccurTwoTimes = getStringsThatOccurNTimes(stringsList, 2);
List<String> stringsThatOccurThreeTimes = getStringsThatOccurNTimes(stringsList, 3);
List<String> stringsThatOccurFourTimes = getStringsThatOccurNTimes(stringsList, 4);
System.out.println("Original list: " + stringsList);
System.out.println("Strings that occur two times: " + stringsThatOccurTwoTimes);
System.out.println("Strings that occur three times: " + stringsThatOccurThreeTimes);
System.out.println("Strings that occur four times: " + stringsThatOccurFourTimes);
}
}
Output:
Original list: [a, b, c, d, e, b, c, c, d, d, d, e]
Strings that occur two times: [b, e]
Strings that occur three times: [c]
Strings that occur four times: [d]
answered Jan 2 at 16:35
shash678shash678
5,93521030
edited Jan 4 at 14:54
answered Jan 2 at 16:35
shash678shash678
5,93521030
answered Jan 2 at 16:35
shash678shash678
5,93521030
answered Jan 2 at 16:35
shash678shash678
5,93521030
5,93521030
How about a util for this in general?
– Naman
Jan 2 at 17:13
Nice I use something similar in python when I do interview questions that are similar to OP's one.
– shash678
Jan 2 at 17:28
add a comment |
How about a util for this in general?
– Naman
Jan 2 at 17:13
Nice I use something similar in python when I do interview questions that are similar to OP's one.
– shash678
Jan 2 at 17:28
How about a util for this in general?
– Naman
Jan 2 at 17:13
How about a util for this in general?
– Naman
Jan 2 at 17:13
Nice I use something similar in python when I do interview questions that are similar to OP's one.
– shash678
Jan 2 at 17:28
Nice I use something similar in python when I do interview questions that are similar to OP's one.
– shash678
Jan 2 at 17:28
add a comment |
The best way to do this is.....
Making of required arraylist.........
ArrayList<String> thrice=new ArrayList<>();
ArrayList<String> one=new ArrayList<>();
Adding some values for checking.......
one.add("1");one.add("1");one.add("1");one.add("1");one.add("1");
one.add("2");one.add("2");one.add("2");one.add("2");one.add("2");
one.add("1");one.add("1");one.add("1");
one.add("3");one.add("3");
one.add("4");one.add("5");
one.add("2");one.add("2");one.add("2");
Mapping to done the task.....
Map<String, Integer> duplicates = new HashMap<String, Integer>();
for (String str : one) {
if (duplicates.containsKey(str)) {
duplicates.put(str, duplicates.get(str) + 1);
} else {
duplicates.put(str, 1);
}
}
for (Map.Entry<String, Integer> entry : duplicates.entrySet()) {
if(entry.getValue()>=3){
thrice.add(entry.getKey());
}
}
Set the list in listview...
ArrayAdapter<String> ad=new ArrayAdapter<>
(this,android.R.layout.simple_list_item_1,thrice);
ListView lv=findViewById(R.id.new_l);
lv.setAdapter(ad);
Output= 1,2
Already check the answer and you can change the condition according to your way
if(entry.getValue()>=3){
thrice.add(entry.getKey());
}
Easiest answer from above...not required to change the versions of minimum sdk for android user
answered Jan 5 at 7:10
Vipul ChauhanVipul Chauhan
4310
add a comment |
The best way to do this is.....
Making of required arraylist.........
ArrayList<String> thrice=new ArrayList<>();
ArrayList<String> one=new ArrayList<>();
Adding some values for checking.......
one.add("1");one.add("1");one.add("1");one.add("1");one.add("1");
one.add("2");one.add("2");one.add("2");one.add("2");one.add("2");
one.add("1");one.add("1");one.add("1");
one.add("3");one.add("3");
one.add("4");one.add("5");
one.add("2");one.add("2");one.add("2");
Mapping to done the task.....
Map<String, Integer> duplicates = new HashMap<String, Integer>();
for (String str : one) {
if (duplicates.containsKey(str)) {
duplicates.put(str, duplicates.get(str) + 1);
} else {
duplicates.put(str, 1);
}
}
for (Map.Entry<String, Integer> entry : duplicates.entrySet()) {
if(entry.getValue()>=3){
thrice.add(entry.getKey());
}
}
Set the list in listview...
ArrayAdapter<String> ad=new ArrayAdapter<>
(this,android.R.layout.simple_list_item_1,thrice);
ListView lv=findViewById(R.id.new_l);
lv.setAdapter(ad);
Output= 1,2
Already check the answer and you can change the condition according to your way
if(entry.getValue()>=3){
thrice.add(entry.getKey());
}
Easiest answer from above...not required to change the versions of minimum sdk for android user
answered Jan 5 at 7:10
Vipul ChauhanVipul Chauhan
4310
add a comment |
The best way to do this is.....
Making of required arraylist.........
ArrayList<String> thrice=new ArrayList<>();
ArrayList<String> one=new ArrayList<>();
Adding some values for checking.......
one.add("1");one.add("1");one.add("1");one.add("1");one.add("1");
one.add("2");one.add("2");one.add("2");one.add("2");one.add("2");
one.add("1");one.add("1");one.add("1");
one.add("3");one.add("3");
one.add("4");one.add("5");
one.add("2");one.add("2");one.add("2");
Mapping to done the task.....
Map<String, Integer> duplicates = new HashMap<String, Integer>();
for (String str : one) {
if (duplicates.containsKey(str)) {
duplicates.put(str, duplicates.get(str) + 1);
} else {
duplicates.put(str, 1);
}
}
for (Map.Entry<String, Integer> entry : duplicates.entrySet()) {
if(entry.getValue()>=3){
thrice.add(entry.getKey());
}
}
Set the list in listview...
ArrayAdapter<String> ad=new ArrayAdapter<>
(this,android.R.layout.simple_list_item_1,thrice);
ListView lv=findViewById(R.id.new_l);
lv.setAdapter(ad);
Output= 1,2
Already check the answer and you can change the condition according to your way
if(entry.getValue()>=3){
thrice.add(entry.getKey());
}
Easiest answer from above...not required to change the versions of minimum sdk for android user
answered Jan 5 at 7:10
Vipul ChauhanVipul Chauhan
4310
The best way to do this is.....
Making of required arraylist.........
ArrayList<String> thrice=new ArrayList<>();
ArrayList<String> one=new ArrayList<>();
Adding some values for checking.......
one.add("1");one.add("1");one.add("1");one.add("1");one.add("1");
one.add("2");one.add("2");one.add("2");one.add("2");one.add("2");
one.add("1");one.add("1");one.add("1");
one.add("3");one.add("3");
one.add("4");one.add("5");
one.add("2");one.add("2");one.add("2");
Mapping to done the task.....
Map<String, Integer> duplicates = new HashMap<String, Integer>();
for (String str : one) {
if (duplicates.containsKey(str)) {
duplicates.put(str, duplicates.get(str) + 1);
} else {
duplicates.put(str, 1);
}
}
for (Map.Entry<String, Integer> entry : duplicates.entrySet()) {
if(entry.getValue()>=3){
thrice.add(entry.getKey());
}
}
Set the list in listview...
ArrayAdapter<String> ad=new ArrayAdapter<>
(this,android.R.layout.simple_list_item_1,thrice);
ListView lv=findViewById(R.id.new_l);
lv.setAdapter(ad);
Output= 1,2
Already check the answer and you can change the condition according to your way
if(entry.getValue()>=3){
thrice.add(entry.getKey());
}
Easiest answer from above...not required to change the versions of minimum sdk for android user
answered Jan 5 at 7:10
Vipul ChauhanVipul Chauhan
4310
answered Jan 5 at 7:10
Vipul ChauhanVipul Chauhan
4310
answered Jan 5 at 7:10
Vipul ChauhanVipul Chauhan
4310
answered Jan 5 at 7:10
Vipul ChauhanVipul Chauhan
4310
4310
add a comment |
add a comment |
You can use computeIfAbsent for this.
List<String> strings;
final Map<String, BigInteger> stringCounts = new HashMap<>();
strings.stream().forEach(s -> {
BigInteger count = stringCounts.computeIfAbsent(s, k -> BigInteger.ZERO).add(BigInteger.ONE);
stringCounts.put(s, count);
});
Now filter stringCounts with value=3.
answered Jan 2 at 17:06
fastcodejavafastcodejava
24.6k19109162
add a comment |
You can use computeIfAbsent for this.
List<String> strings;
final Map<String, BigInteger> stringCounts = new HashMap<>();
strings.stream().forEach(s -> {
BigInteger count = stringCounts.computeIfAbsent(s, k -> BigInteger.ZERO).add(BigInteger.ONE);
stringCounts.put(s, count);
});
Now filter stringCounts with value=3.
answered Jan 2 at 17:06
fastcodejavafastcodejava
24.6k19109162
add a comment |
You can use computeIfAbsent for this.
List<String> strings;
final Map<String, BigInteger> stringCounts = new HashMap<>();
strings.stream().forEach(s -> {
BigInteger count = stringCounts.computeIfAbsent(s, k -> BigInteger.ZERO).add(BigInteger.ONE);
stringCounts.put(s, count);
});
Now filter stringCounts with value=3.
answered Jan 2 at 17:06
fastcodejavafastcodejava
24.6k19109162
You can use computeIfAbsent for this.
List<String> strings;
final Map<String, BigInteger> stringCounts = new HashMap<>();
strings.stream().forEach(s -> {
BigInteger count = stringCounts.computeIfAbsent(s, k -> BigInteger.ZERO).add(BigInteger.ONE);
stringCounts.put(s, count);
});
Now filter stringCounts with value=3.
answered Jan 2 at 17:06
fastcodejavafastcodejava
24.6k19109162
answered Jan 2 at 17:06
fastcodejavafastcodejava
24.6k19109162
answered Jan 2 at 17:06
fastcodejavafastcodejava
24.6k19109162
answered Jan 2 at 17:06
fastcodejavafastcodejava
24.6k19109162
24.6k19109162
add a comment |
add a comment |
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1
Basically, instead of using a HashSet you now want to use a HashMap to count the number of occurrences of each string.
– shash678
Jan 2 at 16:31
See this answer for a solution as @shash678 suggested
– Stalemate Of Tuning
Jan 2 at 16:35
Possible duplicate of get the duplicates values from Arraylist<String> and then get those items in another Arraylist
– Vipul Chauhan
Jan 4 at 17:25