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

• Difficulty Level : Medium
• Last Updated : 23 Mar, 2022

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 ``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

 ``

Output

`2`

Efficient Approach: Follow the steps below to solve the problem:

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ``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 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 v = ``new` `ArrayList<>();` `    ``// Traverse the map``    ``for` `(Map.Entry 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

 ``

Output

`2`

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

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