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  1. Counting Sort 1
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Counting Sort 1

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690 Discussions

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  • somyarajsinghda1
     + 0 comments

    why its not working for 5th test case def countingSort(arr): arr.sort() d=arr[-1] l=[] for i in range(d+1): count=0 for j in arr: if i==j: count+=1 l.append(count) return l

  • ataquecardiacob1
     + 0 comments

    C#

    public static List<int> countingSort(List<int> arr)
    {
        var res = new int[100];      
        
        foreach (var item in arr) {
            res[item] += 1;
        }
        
        return res.ToList();
    }

  • akulbachandra67
     + 0 comments

    mine python solution

    def CountingSort(arr):
    lps = [0]*(max(arr)+1)
    for i in arr:
        lps[i] += 1
    return lps

    correct solution according to hacker rank

    def CountingSort(arr):
    lps = [0]*(100)
    for i in arr:
        lps[i] += 1
    return lps

  • mdsanz
     + 0 comments

    My Java 8 Solution:

    public static List<Integer> countingSort(List<Integer> arr) {
        List<Integer> result = new ArrayList<>(Collections.nCopies(100, 0));
        
        for (int number : arr) {
            result.set(number, result.get(number) + 1);
        }
        
        return result;
    }

  • tannar_mj
     + 1 comment

    A couple Python solutions:

    Fun one liner:

    def countingSort(arr):
    return [arr.count(i) for i in range(100)]

    Not very optimal, O(n^2) time complexity

    Better but more verbose solution for O(n) time complexity:

    def countingSort(arr):
    counts = [0 for i in range(100)]
    for n in arr:
        counts[n] += 1
    return [counts[i] for i in range(100)]