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

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

Naive Approach: We count the frequency of each element in an array. Then for each value v in frequency map, we traverse the frequency map and check whether this current value is less than v, if it is true then we add this current value to our result and if it is false then we add the difference between the current value and v to our result. After each traversal, store minimum of current result and previous result.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to get minimum removals required
// to make frequency of all remaining elements equal
int minDelete(int arr[], int n)
{
    // Create an hash map and store frequencies of all
    // array elements in it using element as key and
    // frequency as value
    unordered_map<int, int> freq;
    for (int i = 0; i < n; i++)
        freq[arr[i]]++;

    // Initialize the result to store the minimum deletions
    int tempans, res = INT_MAX;

    // Find deletions required for each element and store
    // the minimum deletions in result
    for (auto itr = freq.begin(); itr != freq.end();
         itr++) {
        tempans = 0;
        for (auto j = freq.begin(); j != freq.end(); j++) {
            if (j->second < itr->second) {
                tempans = tempans + j->second;
            }
            else {
                tempans
                    = tempans + (j->second - itr->second);
            }
        }
        res = min(res, tempans);
    }

    return res;
}

// Driver program to run the case
int main()
{
    int arr[] = {2, 4, 3, 2, 5, 3};
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << minDelete(arr, n);
    return 0;
}

Python3

# Python program for the above approach

# Function to get minimum removals required
# to make frequency of all remaining elements equal
import sys

def minDelete(arr, n):

    # Create an hash map and store frequencies of all
    # array elements in it using element as key and
    # frequency as value
    freq = {}
    for i in range(n):
        if(arr[i] in freq):
            freq[arr[i]] = freq[arr[i]]+1
        else:
            freq[arr[i]] = 1

    # Initialize the result to store the minimum deletions
    tempans, res = sys.maxsize,sys.maxsize

    # Find deletions required for each element and store
    # the minimum deletions in result
    for [key,value] in freq.items():
        tempans = 0
        for [key1,value1] in freq.items():
            if (value1 < value):
                tempans = tempans + value1
            else:
                tempans = tempans + (value1 - value)

        res = min(res, tempans)

    return res

# Driver program to run the case
arr = [2, 4, 3, 2, 5, 3]
n = len(arr)
print(minDelete(arr, n))

# This code is contributed by shinjanpatra.

Javascript

<script>

// JavaScript program for the above approach

// Function to get minimum removals required
// to make frequency of all remaining elements equal
function minDelete(arr, n)
{
    // Create an hash map and store frequencies of all
    // array elements in it using element as key and
    // frequency as value
    let freq = new Map();
    for (let i = 0; i < n; i++){
        if(freq.has(arr[i])){
            freq.set(arr[i],freq.get(arr[i])+1);
        }
        else freq.set(arr[i],1);
    }

    // Initialize the result to store the minimum deletions
    let tempans, res = Number.MAX_VALUE;

    // Find deletions required for each element and store
    // the minimum deletions in result
    for (let [key,value] of freq){
        let tempans = 0;
        for (let [key1,value1] of freq) {
            if (value1 < value) {
                tempans = tempans + value1;
            }
            else {
                tempans = tempans + (value1 - value);
            }
        }
        res = Math.min(res, tempans);
    }

    return res;
}

// Driver program to run the case
let arr = [2, 4, 3, 2, 5, 3];
let n = arr.length;
document.write(minDelete(arr, n));

// This code is contributed by shinjanpatra.
</script>

Output

2

Efficient 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);
}

Java

/*package whatever //do not write package name here */

import java.io.*;
import java.util.HashMap;
import java.util.Map;
import java.util.ArrayList;
import java.util.Collections;

class GFG {
  public static void minDeletions(int arr[], int N)
  {

    // Stores frequency of
    // all array elements
    HashMap<Integer, Integer> map = new HashMap<>();
    ;

    // Traverse the array
    for (int i = 0; i < N; i++) {
      Integer k = map.get(arr[i]);
      map.put(arr[i], (k == null) ? 1 : k + 1);
    }

    // Stores all the frequencies
    ArrayList<Integer> v = new ArrayList<>();

    // Traverse the map
    for (Map.Entry<Integer, Integer> e :
         map.entrySet()) {
      v.add(e.getValue());
    }

    // Sort the frequencies
    Collections.sort(v);

    // Count of frequencies
    int size = v.size();

    // Stores the final count
    int ans = N - (v.get(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.get(i) != v.get(i - 1)) {
        int safe = v.get(i) * (size - i);
        ans = Math.min(ans, N - safe);
      }
    }

    // Print the final count
    System.out.println(ans);
  }

  // Driver code
  public static void main(String[] args)
  {

    // Given array
    int arr[] = { 2, 4, 3, 2, 5, 3 };

    // Size of the array
    int N = 6;

    // Function call to print the minimum
    // number of removals required
    minDeletions(arr, N);
  }
}

// This code is contributed by aditya7409.

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.

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{

  // Function to count the minimum
  // removals required to make frequency
  // of all array elements equal
  static void minDeletions(int []arr, int N)
  {

    // Stores frequency of
    // all array elements
    Dictionary<int,int> freq = new Dictionary<int,int>();

    // Traverse the array
    for (int i = 0; i < N; i++)
    {
      if(freq.ContainsKey(arr[i]))
        freq[arr[i]]++;
      else
        freq[arr[i]] = 1;
    }

    // Stores all the frequencies
    List<int> v = new List<int>();

    // Traverse the map
    foreach (var z in freq) {
      v.Add(z.Value);
    }

    // Sort the frequencies
    int sz = v.Count;
    int []temp = new int[sz];
    for(int i = 0; i < v.Count; i++)
      temp[i] = v[i];
    Array.Sort(temp);
    for(int i = 0; i < v.Count; i++)
      v[i] = temp[i];

    // Count of frequencies
    int size = v.Count;

    // Stores the final count
    int ans = N - (v[0] * size);

    // Traverse the vector
    for (int i = 1; i < v.Count; 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 = Math.Min(ans, N - safe);
      }
    }

    // Print the final count
    Console.WriteLine(ans);
  }

  // Driver Code
  public static void Main()
  {

    // Given array
    int []arr = { 2, 4, 3, 2, 5, 3 };

    // Size of the array
    int N = arr.Length;

    // Function call to print the minimum
    // number of removals required
    minDeletions(arr, N);
  }
}

// This code is contributed by SURENDRA_GANGWAR.

Javascript

<script>

// Javascript program for the above approach

// Function to count the minimum
// removals required to make frequency
// of all array elements equal
function minDeletions(arr, N)
{
    // Stores frequency of
    // all array elements
    var freq = new Map();

    // Traverse the array
    for (var i = 0; i < N; i++) {
        if(freq.has(arr[i]))
        {
            freq.set(arr[i], freq.get(arr[i])+1);
        }
        else
        {
            freq.set(arr[i], 1);
        }
    }

    // Stores all the frequencies
    var v = [];

    // Traverse the map
    freq.forEach((value, key) => {
        v.push(value);
    });

    // Sort the frequencies
    v.sort();

    // Count of frequencies
    var size = v.length;

    // Stores the final count
    var ans = N - (v[0] * size);

    // Traverse the vector
    for (var i = 1; i < v.length; i++) {

        // Count the number of removals
        // for each frequency and update
        // the minimum removals required
        if (v[i] != v[i - 1]) {
            var safe = v[i] * (size - i);
            ans = Math.min(ans, N - safe);
        }
    }

    // Print the final count
    document.write( ans);
}

// Driver Code
// Given array
var arr = [ 2, 4, 3, 2, 5, 3 ];
// Size of the array
var N = arr.length;
// Function call to print the minimum
// number of removals required
minDeletions(arr, N);

</script>

Output

2

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