Minimum removals required to make frequency of all remaining array elements equal

Last Updated : 25 Feb, 2021

Given an array arr[] of size N, the task is to find the minimum number of array elements required to be removed such that the frequency of the remaining array elements become equal.

Examples :

Input: arr[] = {2, 4, 3, 2, 5, 3}
Output: 2
Explanation: Following two possibilities exists:
1) Either remove an occurrence of 2 and 3. The array arr[] modifies to {2, 4, 3, 5}. Therefore, frequency of all the elements become equal.
2) Or, remove an occurrence of 4 and 5. The array arr[] modifies to {2, 3, 2, 3}. Therefore, frequency of all the elements become equal.

Input: arr[] = {1, 1, 2, 1}
Output: 1

Approach: Follow the steps below to solve the problem:

Initialize an ordered map, say freq, to store the frequency of all array elements.
Traverse the array arr[] and store the respective frequencies of array elements.
Initialize a vector, say v, to store all the frequencies stored in the map.
Sort the vector v.
Initialize a variable, say ans, to store the final count.
Traverse the vector v and for each frequency, count the number of elements required to be removed to make the frequency of the remaining elements equal.
Print the minimum of all the count of removals obtained.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to count the minimum
// removals required to make frequency
// of all array elements equal
int minDeletions(int arr[], int N)
{
    // Stores frequency of
    // all array elements
    map<int, int> freq;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
        freq[arr[i]]++;
    }
 
    // Stores all the frequencies
    vector<int> v;
 
    // Traverse the map
    for (auto z : freq) {
        v.push_back(z.second);
    }
 
    // Sort the frequencies
    sort(v.begin(), v.end());
 
    // Count of frequencies
    int size = v.size();
 
    // Stores the final count
    int ans = N - (v[0] * size);
 
    // Traverse the vector
    for (int i = 1; i < v.size(); i++) {
 
        // Count the number of removals
        // for each frequency and update
        // the minimum removals required
        if (v[i] != v[i - 1]) {
            int safe = v[i] * (size - i);
            ans = min(ans, N - safe);
        }
    }
 
    // Print the final count
    cout << ans;
}
 
// Driver Code
int main()
{
    // Given array
    int arr[] = { 2, 4, 3, 2, 5, 3 };
 
    // Size of the array
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function call to print the minimum
    // number of removals required
    minDeletions(arr, N);
}

Python3

# Python 3 program for the above approach
from collections import defaultdict
 
# Function to count the minimum
# removals required to make frequency
# of all array elements equal
def minDeletions(arr, N):
   
    # Stores frequency of
    # all array elements
    freq = defaultdict(int)
 
    # Traverse the array
    for i in range(N):
        freq[arr[i]] += 1
 
    # Stores all the frequencies
    v = []
 
    # Traverse the map
    for z in freq.keys():
        v.append(freq[z])
 
    # Sort the frequencies
    v.sort()
 
    # Count of frequencies
    size = len(v)
 
    # Stores the final count
    ans = N - (v[0] * size)
 
    # Traverse the vector
    for i in range(1, len(v)):
 
        # Count the number of removals
        # for each frequency and update
        # the minimum removals required
        if (v[i] != v[i - 1]):
            safe = v[i] * (size - i)
            ans = min(ans, N - safe)
 
    # Print the final count
    print(ans)
 
 
# Driver Code
if __name__ == "__main__":
 
    # Given array
    arr = [2, 4, 3, 2, 5, 3]
 
    # Size of the array
    N = len(arr)
 
    # Function call to print the minimum
    # number of removals required
    minDeletions(arr, N)
 
    # This code is contributed by chitranayal.

Output:

Time Complexity: O(N)
Auxiliary Space: O(N)

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