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  1. Counting Sort 1
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Counting Sort 1

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  • crono7
     + 0 comments

    Java

    List<Integer> frequencies = new ArrayList<>();
for(int i =0; i < 100; i++){
		frequencies.add(Collections.frequency(arr, i));
}
return frequencies;

  • rosezhaott
     + 0 comments

    Java 8:

    • public static List countingSort(List arr) { Map countMap = new HashMap<>(); List result = new ArrayList<>(); for (int i = 0; i< 100; i++) { countMap.put(i, 0); }

      for (int num : arr) { int value = countMap.get(num); countMap.put(num, value + 1); }

      for (int i = 0; i < 100; i++) { result.add(countMap.get(i)); }

      return result;

      }

  • fspng1
     + 0 comments

    In the question description, the input format is as follows:

    Input Format

    The first line contains an integer , the number of items in . Each of the next lines contains an integer where .

    If you solve it with C++, its input is a vector and the size() tells you what N is.

    So there is no first line input.

  • kumarsahudeepak1
     + 0 comments

    Scala solution

    def countingSort(arr: Array[Int]): Array[Int] = {
    // Write your code here
        val frequency = Array.fill(100)(0)
        for (num <- arr) {
        frequency(num) += 1
        }
        frequency
    }

  • chauhanaman00000
     + 0 comments
     int totalCount = arr.Count;

 List<int> a = new List<int>(totalCount);
 for(int i =0;i< 100;i++)
 {
     a.Add(0);
 }
 int count = 0;
 for (int i = 0; i < arr.Count; i++)
 {
     if (a[arr[i]] > 0)
     {
         a[arr[i]] =++a[arr[i]];
     }
     else
     {
         a[arr[i]] = ++count;
     }
     count = 0;
 }