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

Counting Sort 2

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396 Discussions

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  • afzal_saiyed931
     + 0 comments
    def countingSort(arr):
    # Write your code here
    
    new_arr = [0] * 100
    
    for i in arr:
        new_arr[i] += 1
    sorted_arr = []  
    for i in range(len(new_arr)):
        for j in range(new_arr[i]):
            sorted_arr.append(i)
            
    return sorted_arr

  • 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());
    }