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  1. Prepare
  2. Algorithms
  3. Sorting
  4. Counting Sort 1
  5. Discussions

Counting Sort 1

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465 Discussions

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  • alban_tyrex
     + 0 comments

    Here is my c++ solution, you can watch the explanation here : https://youtu.be/snADwNvEEcM

    vector<int> countingSort(vector<int> arr) {
    vector<int> count(100, 0);
    for(int i = 0; i < arr.size(); i++) count[arr[i]]++;
    return count;
}

  • emhhacker
     + 0 comments

    My Java solution with o(n) time complexity and o(1) space complexity:

    public static List<Integer> countingSort(List<Integer> arr) {
        // init freq array
        int[] freqArr = new int[100];
        
        //update freq for each val
        for(int element : arr) freqArr[element]++;
        
        //return int array converted into Integer list
        return Arrays.stream(freqArr)
            .boxed()
            .collect(Collectors.toList());
    }

  • patrickgao2020
     + 0 comments

    Here is my simple Python solution!

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

  • kaadir_yaalcin
     + 0 comments
    public static List<int> countingSort(List<int> arr)
{
    var countMap = arr.GroupBy(n => n)
                    .OrderBy(g => g.Key)
                    .ToDictionary(g => g.Key, g => g.Count());

    List<int> zeros = Enumerable.Repeat(0, 100).ToList();

    for (int i = 0; i < zeros.Count; i++)
    {
        if (countMap.ContainsKey(i))
        {
            zeros[i] = countMap.GetValueOrDefault(i); 
        }
    }

    return zeros;
}

  • vuhuutuan0
     + 0 comments

    My answer in typescript, simple

    function countingSort(n: number, arr: number[]): number[] {
    let arr_count = Array(100).fill(0)

    for (let i = 0; i < arr.length; i++) arr_count[arr[i]]++

    return arr_count;
}