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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Equalize the Array
  5. Discussions

Equalize the Array

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1930 Discussions

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  • Arkeonn_
     + 0 comments

    JS solution :

    function equalizeArray(arr) {
    return arr.length - Math.max(...[...new Set(arr)].map((num, i) => arr.reduce((acc, next) => next == num ? acc + 1 : acc, 0)))
}

  • alban_tyrex
     + 0 comments

    Here is my O(N) c++ solution, you can watch the explanation here : https://youtu.be/8p9yuqSv4ek

    int equalizeArray(vector<int> arr) {
    map<int, int>mp;
    int mx = 0;
    for(int el: arr) {
        mp[el]++;
        if(mp[el]>mx) mx = mp[el];
    }
    return arr.size() - mx;
}

  • pt023150
     + 1 comment

    java 8

    public static int equalizeArray(List arr) { int max = 0; // Write your code here

        HashMap<Integer, Integer> map = new HashMap<>();
    for (int el : arr) {
        if (map.containsKey(el)) {
            map.put(el, map.get(el) + 1);
        } else {
            map.put(el, 1);
        }

    }

    Set<Integer> keys = map.keySet();
    for (int key : keys) {
        if (map.get(key) > max) {
            max = map.get(key);
        }

    }

    return arr.size() - max;

}

  • jmuchaves
     + 0 comments

    Javascript:

    function equalizeArray(arr) {
    const counts = {};
    let max = 0;
    for (let i = 0; i < arr.length; i++) {
        counts[arr[i]] = (counts[arr[i]] || 0) + 1;
        max = Math.max(counts[arr[i]], max);
    }
    return arr.length - max;
}

  • nischayd75
     + 0 comments
      public static int equalizeArray(List<Integer> arr) {
    HashMap<Integer, Integer> ele = new HashMap<>();
    for(int n : arr)
       ele.put(n, Collections.frequency(arr, n));
    return arr.size() - Collections.max(ele.values());  
    }