# Minimum operation to make all elements equal in array

Last Updated : 17 Apr, 2023

Given an array with n positive integers. We need to find the minimum number of operations to make all elements equal. We can perform addition, multiplication, subtraction, or division with any part on an array element.

Examples:

```Input : arr[] = {1, 2, 3, 4}
Output : 3
Since all elements are different,
we need to perform at least three
operations to make them same. For
example, we can make them all 1
by doing three subtractions. Or make
them all 3 by doing three additions.

Input : arr[] = {1, 1, 1, 1}
Output : 0```

To make all elements equal you can select a target value and then you can make all elements equal to that. Now, for converting a single element to target value you can perform a single operation only once. In this manner, you can achieve your task in a maximum of n operations but you have to minimize this number of operations and for this, your selection of target is very important because if you select a target whose frequency in array is x then you have to perform only n-x more operations as you have already x elements equal to your target value. So finally, our task is reduced to finding the element with maximum frequency. This can be achieved by different means such as the iterative method in O(n^2), sorting in O(nlogn), and hashing in O(n) time complexity.

Step-by-step approach:

• Create an empty hash table to store the frequency of each element in the array.
• Traverse the array and insert each element into the hash table. If an element is already present in the hash table, increment its frequency.
• Find the maximum frequency of any element in the hash table.
• The minimum number of operations required to make all elements equal is equal to the difference between the total number of elements in the array and the maximum frequency of any element in the hash table

Pseudocode:

```minOperations(arr, n)
// Step 1
hashTable = {}
// Step 2
for i = 0 to n-1 do
if arr[i] is in hashTable then
hashTable[arr[i]] = hashTable[arr[i]] + 1
else
hashTable[arr[i]] = 1
end if
end for
// Step 3
maxCount = 0
for key in hashTable do
if hashTable[key] > maxCount then
maxCount = hashTable[key]
end if
end for
// Step 4
return n - maxCount
end function```

Implementation:

## C++

 `// CPP program to find the minimum number of ` `// operations required to make all elements ` `// of array equal` `#include ` `#include ` `#include `   `using` `namespace` `std;`   `int` `minOperation(``int` `arr[], ``int` `n) {` `    ``// Insert all elements in hash.` `    ``unordered_map<``int``, ``int``> hash;` `    ``for` `(``int` `i=0; i

## Java

 `// JAVA Code For Minimum operation to make` `// all elements equal in array` `import` `java.util.*;`   `class` `GFG {` `    `  `    ``// function for min operation ` `    ``public` `static` `int` `minOperation (``int` `arr[], ``int` `n) ` `    ``{` `        ``// Insert all elements in hash. ` `       ``HashMap hash = ``new` `HashMap(); ` `        `  `        ``for` `(``int` `i=``0``; i s = hash.keySet(); ` `        `  `        ``for` `(``int` `i : s)` `            ``if` `(max_count < hash.get(i))` `                ``max_count = hash.get(i);` `     `  `        ``// return result` `        ``return` `(n - max_count); ` `    ``}` `    `  `    ``/* Driver program to test above function */` `    ``public` `static` `void` `main(String[] args) ` `    ``{` `        ``int` `arr[] = {``1``, ``5``, ``2``, ``1``, ``3``, ``2``, ``1``};` `        ``int` `n = arr.length;` `        ``System.out.print(minOperation(arr, n));` `            `  `    ``}` `}` `  `  `// This code is contributed by Arnav Kr. Mandal.`

## Python3

 `# Python3 program to find the minimum ` `# number of operations required to ` `# make all elements of array equal ` `#from collections import defaultdict`   `# Function for min operation ` `# using hashing` `def` `minOperation(arr, n):` `    ``mp ``=` `{}` `    ``max_freq ``=` `0` `    ``#update the map` `    ``for` `i ``in` `arr:` `        ``mp[i] ``=` `mp.get(i,``0``) ``+` `1` `    ``#finding max freq ` `    ``for` `i ``in` `mp:` `        ``if` `mp[i] > max_freq:` `            ``max_freq ``=` `mp[i]` `    ``# if all elements have same freq we have to change n-1 elements` `    ``if` `max_freq ``=``=` `1``:` `        ``return` `n``-``1` `    ``# we need to change only remaining elements ` `    ``return` `(n``-``max_freq)` `    `  `        `  `      `  `# Driver Code` `if` `__name__ ``=``=` `"__main__"``: `   `    ``arr ``=` `[``1``, ``5``, ``2``, ``1``, ``3``, ``2``, ``1``] ` `    ``n ``=` `len``(arr) ` `    ``print``(minOperation(arr, n))` `    `  `# This code is contributed ` `# by Aditya Bharat`

## C#

 `// C# Code For Minimum operation to make` `// all elements equal in array` `using` `System;` `using` `System.Collections.Generic;` `    `  `class` `GFG ` `{` `    `  `    ``// function for min operation ` `    ``public` `static` `int` `minOperation (``int` `[]arr, ``int` `n) ` `    ``{` `        ``// Insert all elements in hash. ` `        ``Dictionary<``int``,``int``> m = ``new` `Dictionary<``int``,``int``>();` `        ``for` `(``int` `i = 0 ; i < n; i++)` `        ``{` `            ``if``(m.ContainsKey(arr[i]))` `            ``{` `                ``var` `val = m[arr[i]];` `                ``m.Remove(arr[i]);` `                ``m.Add(arr[i], val + 1); ` `            ``}` `            ``else` `            ``{` `                ``m.Add(arr[i], 1);` `            ``}` `        ``}` `        `  `        ``// find the max frequency` `        ``int` `max_count = 0;` `        ``HashSet<``int``> s = ``new` `HashSet<``int``>(m.Keys); ` `        `  `        ``foreach` `(``int` `i ``in` `s)` `            ``if` `(max_count < m[i])` `                ``max_count = m[i];` `    `  `        ``// return result` `        ``return` `(n - max_count); ` `    ``}` `    `  `    ``/* Driver code */` `    ``public` `static` `void` `Main(String[] args) ` `    ``{` `        ``int` `[]arr = {1, 5, 2, 1, 3, 2, 1};` `        ``int` `n = arr.Length;` `        ``Console.Write(minOperation(arr, n));` `            `  `    ``}` `}`   `// This code is contributed by 29AjayKumar`

## Javascript

 ``

Output

`4`

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