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# 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.

Algorithm:

Step 1: Create a function called “minDelete” that accepts the arguments arr, an integer array of size n, and the function name.
Step 2: Create an unordered map called “freq” to keep track of each element’s frequency in the input array. Set each element’s                    frequency to 0 at the beginning.
Step 3: Increase the frequency of each element in the “freq” map as you traverse the input array.
Step 4: Set two variables, “tempans” and “res,” to their respective initial values of 0 and INT MAX.
Step 5: Use the iterator “itr” to navigate the “freq” map. Repeat the traversal of the map for each element using an additional                      iterator called “j”.
a. Increase “tempans” by the frequency of the element pointed by “j” if its frequency is lower than that of the element                          pointed by “itr”.
b. If the frequency of the element pointed by “j” is greater than or equal to that of the element pointed by “itr”, increment                   “tempans” by the difference between the frequency of the element pointed by “j” and that of the element pointed by                     “itr”.
Step 6: Update “res” with the minimum value between “res” and “tempans”.
Step 7: Return res.

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);` `    ``cout << minDelete(arr, n);` `    ``return` `0;` `}`

## Java

 `import` `java.util.HashMap;` `import` `java.util.Map;`   `// Java program for the above approach` `class` `Main ` `{` `  `  `  ``// Function to get minimum removals required` `  ``// to make frequency of all remaining elements equal` `  ``static` `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` `    ``Map freq = ``new` `HashMap<>();` `    ``for` `(``int` `i = ``0``; i < n; i++)` `      ``freq.put(arr[i], freq.getOrDefault(arr[i], ``0``) + ``1``);`   `    ``// Initialize the result to store the minimum deletions` `    ``int` `tempans, res = Integer.MAX_VALUE;`   `    ``// Find deletions required for each element and store` `    ``// the minimum deletions in result` `    ``for` `(Map.Entry itr : freq.entrySet()) {` `      ``tempans = ``0``;` `      ``for` `(Map.Entry j : freq.entrySet()) {` `        ``if` `(j.getValue() < itr.getValue()) {` `          ``tempans = tempans + j.getValue();` `        ``}` `        ``else` `{` `          ``tempans = tempans + (j.getValue() - itr.getValue());` `        ``}` `      ``}` `      ``res = Math.min(res, tempans);` `    ``}`   `    ``return` `res;` `  ``}`   `  ``// Driver program to run the case` `  ``public` `static` `void` `main(String args[])` `  ``{` `    ``int` `arr[] = { ``2``, ``4``, ``3``, ``2``, ``5``, ``3` `};` `    ``int` `n = arr.length;` `    ``System.out.println(minDelete(arr, n));` `  ``}` `}`   `// This code is contributed by phasing17.`

## 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.`

## C#

 `// C# program for the above approach` `using` `System;` `using` `System.Collections.Generic;`   `class` `GFG` `{`   `  ``// Function to get minimum removals required` `  ``// to make frequency of all remaining elements equal` `  ``static` `int` `MinDelete(``int``[] arr, ``int` `n)` `  ``{`   `    ``// Create a dictionary and store frequencies of all` `    ``// array elements in it using element as key and` `    ``// frequency as value` `    ``Dictionary<``int``, ``int``> freq = ``new` `Dictionary<``int``, ``int``>();` `    ``for` `(``int` `i = 0; i < n; i++)` `    ``{` `      ``if` `(freq.ContainsKey(arr[i]))` `      ``{` `        ``freq[arr[i]]++;` `      ``}` `      ``else` `      ``{` `        ``freq[arr[i]] = 1;` `      ``}` `    ``}`   `    ``// Initialize the result to store the minimum deletions` `    ``int` `tempans, res = ``int``.MaxValue;`   `    ``// Find deletions required for each element and store` `    ``// the minimum deletions in result` `    ``foreach` `(KeyValuePair<``int``, ``int``> itr ``in` `freq)` `    ``{` `      ``tempans = 0;` `      ``foreach` `(KeyValuePair<``int``, ``int``> j ``in` `freq)` `      ``{` `        ``if` `(j.Value < itr.Value)` `        ``{` `          ``tempans = tempans + j.Value;` `        ``}` `        ``else` `        ``{` `          ``tempans = tempans + (j.Value - itr.Value);` `        ``}` `      ``}` `      ``res = Math.Min(res, tempans);` `    ``}`   `    ``return` `res;` `  ``}`   `  ``// Driver program to run the case` `  ``static` `void` `Main(``string``[] args)` `  ``{` `    ``int``[] arr = { 2, 4, 3, 2, 5, 3 };` `    ``int` `n = arr.Length;` `    ``Console.WriteLine(MinDelete(arr, n));` `  ``}` `}`   `// This code is contributed by phasing17`

## 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 * 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);`   `    ``// 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 * 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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