Time Complexity:

def equalizeArray(arr:list):
    lenght = len(arr)
    frequency = {}
    for num in arr:
        if num not in frequency:
            frequency[num] = 1
        else:
            frequency[num] += 1
    

    return abs(max(frequency.values()) - lenght)

    Comparator<Long> longComparator = Long::compare;
    Map<Integer, Long> valueMap = arr.stream().collect(Collectors.groupingBy(s -> s, Collectors.counting()));
    List<Long> values = valueMap.values().stream().sorted(longComparator.reversed()).collect(Collectors.toList());
    long mostrepeated = values.get(0);
    long totaloccurences= values.stream().mapToLong(s->s).sum();
    return (int)(totaloccurences - mostrepeated);

Jaav 8 solution

    int maxRepeat = 0;

        for(int i = 0; i < arr.size(); i++){
            int num = arr.get(i);
            int RepeatNum = 0;

            for(int j = 0; j < arr.size(); j++){
                if(arr.get(j) == num){
                    RepeatNum++;
                }
            }
            if(maxRepeat < RepeatNum){
                maxRepeat = RepeatNum;
            }
        }
        return arr.size() - maxRepeat;

Python solution

def equalizeArray(arr):
    arr_dict = Counter(arr)
    max_num_arr = max(arr_dict.values())
    return len(arr)-max_num_arr

Here is problem solution in Python, Java, C++, C and Javascript - https://programmingoneonone.com/hackerrank-equalize-the-array-problem-solution.html