Minimum removals required to make frequency of all remaining array elements equal
Last Updated :
23 Jul, 2025
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 <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;
}
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<Integer, Integer> 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<Integer, Integer> itr : freq.entrySet()) {
tempans = 0;
for (Map.Entry<Integer, Integer> 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.
# 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# 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
<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>
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 <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);
}
/*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.
# 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# 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.
<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>
Time Complexity: O(N)
Auxiliary Space: O(N)
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