We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.
  1. Counting Sort 1
  2. Discussions

Counting Sort 1

Sort by

recency

|

267 Discussions

|

  • cyrus_maj
     + 0 comments

    For those who stumbled into the same issue as me, make sure that you understand what a frequency array is first, and the proper way to create one, otherwise you will bang your head into a wall over this problem. I ran into issues because I didn't realize it is supposed to be size 100, for every single case. I was making it the same size as the size as the input array.

  • matedominguez2
     + 0 comments

    Python solution

    def countingSort(arr):
    freq = [0] * 100
    
    for n in arr:
        freq[n] += 1
        
    return freq

  • easha2658
     + 0 comments
    def countingSort(arr):
    # Write your code here
    frequencyArr = [0] * 100
    
    for a in arr:
        frequencyArr[a] += 1
    
    return frequencyArr

  • derekcommongrou1
     + 1 comment

    I feel like this challenge is a lot easier than the challenges suggested before it, and should probably be one of the first ones given.

  • lo4251054
     + 0 comments

    Java Solution:

        public static List<Integer> countingSort(List<Integer> arr) {
        
        Map<Integer, Integer> countMap = new HashMap<>();
        
        for(int i=0; i< arr.size(); i++){
            int current = arr.get(i);
            if(countMap.containsKey(current)){
                int count = countMap.get(current);
                countMap.put(current, count+1);
            }else{
                countMap.put(current, 1);
            }
        }
        
        List<Integer> list = new ArrayList<>(100);
        for(int i=0; i<100; i++){
            if(countMap.containsKey((i))){
                list.add(countMap.get(i));
            }else{
                list.add(0);
            }
        }
        
        return list;
    }