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

Equalize the Array

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1925 Discussions

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  • 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;
}

  • bawa101001
     + 0 comments

    Python solution with O(n) complexity and without using collections library:

    def equalize_array(arr):
    count = {}
    for el in arr:
        if el in count:
            count[el] += 1
        else:
            count[el] = 1
    # Find the maximum frequency
    max_count = 0
    for value in count.values():
        if value > max_count:
            max_count = value
    # Calculate the number of elements to remove
    return len(arr) - max_count

  • Ramesh_Kumar_S
     + 0 comments
    def equalizeArray(arr):
    d=dict(Counter(arr))
    return len(arr) - max(d.values())

  • quangluan0305201
     + 0 comments

    include

    using namespace std; int test(int n, vector a){ unordered_map v; for(int it : a){ v[it]++; } int max_value = 0; for(auto& pair : v){ max_value = max(max_value, pair.second); } return n - max_value; } int main(){ int n; cin >> n; vector a(n); for(int i = 0; i < n; i++){ cin >> a[i]; } cout << test(n, a) << endl; return 0; }

  • folch_bk2
     + 0 comments

    PHP

    function equalizeArray($arr) {
    $counter = [];
    $max = [];
    foreach ($arr as $el) {
        isset($counter[$el]) ? $counter[$el]++ : $counter[$el] = 1;
        if ($counter[$el] > $max['value']) {
            $max['value'] = $counter[$el];
            $max['index'] = $el;
        }
    }
    
    unset($counter[$max['index']]);
    return array_sum($counter);
}