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

Counting Sort 2

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395 Discussions

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  • ahmedmhmouad41
     + 0 comments

    NodeJS

    function countingSort(arr) {
    let count = new Array(100).fill(0);
    let result=[];
    
    for(let i=0;i<arr.length;i++) count[arr[i]]++;
    for(let i=0;i<100;i++) {
        while(count[i]>0){
            result.push(i)
            count[i]--;
        }
    }
    return result
    
}

  • yashdeora98294
     + 0 comments

    Here is problem solution in python java c++ c and Javascript - https://programmingoneonone.com/hackerrank-counting-sort-2-problem-solution.html

  • alban_tyrex
     + 0 comments

    Here is my c++ solution, you can watch the explanation here : https://youtu.be/TPW8IGrTI8A

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

  • emhhacker
     + 0 comments

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

    public static List<Integer> countingSort(List<Integer> arr) {
        int[] freqArr = new int[100];
        arr.forEach(num -> freqArr[num]++);

        return IntStream.range(0, 100)
            .boxed()
            .flatMap(i -> Collections.nCopies(freqArr[i], i).stream())
            .collect(Collectors.toList());
    }

  • michaelchen0611
     + 0 comments
    1. Time Complexity: O(n)
    2. Space Complexity: O(n)
    int* countingSort(int arr_count, int* arr, int* result_count) {
    static int res_map[100] = {0};
    *result_count = 0;
    for (int i = 0; i<arr_count; i++){
        res_map[arr[i]] ++;
    }
    for (int i = 0; i<100; i++){
        if (res_map[i] != 0){
            for (int j = 0; j<res_map[i]; j++){
                arr[(*result_count)] = i;
                (*result_count)++;
            }
        }
    }
    return arr;
}