# Minimize increments required to make differences between all pairs of array elements even

• Last Updated : 15 Feb, 2022

Given an array, arr[] consisting of N integers, the task is to minimize the number of increments of array elements required to make all differences pairs of array elements even.

Examples:

Input: arr[] = {4, 1, 2}
Output: 1
Explanation:

Operation 1: Increment arr[1] by 1. The array arr[] modifies to {4, 2, 2}.
All pairs: (4, 2) â†’ difference = 2
(4, 2) â†’ difference = 2
(2, 2) â†’ difference = 0
Now, the pairwise differences between array elements is even. Hence, the answer is 1.

Input: arr[] = {2, 4}
Output: 0
Explanation: Differences between all pairs of array elements is already even. Therefore, the answer is 0.

Approach: The given problem can be solved by observing the fact that, to make the difference between all pairs of array elements even, both the elements must be of the same parity. Therefore, the idea is to convert all the array elements to either even numbers or odd numbers. The minimum number of increments is equal to the minimum of the count of even and odd array elements

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to find the minimum increments``// required to difference between all``// pairs of array elements even``void` `minimumOperations(``int` `arr[], ``int` `N)``{``    ``// Store the count of odd``    ``// and even numbers``    ``int` `oddCnt = 0, evenCnt = 0;` `    ``// Traverse the array``    ``for` `(``int` `i = 0; i < N; i++) {` `        ``if` `(arr[i] % 2 == 0) {` `            ``// Increment evenCnt by 1``            ``evenCnt++;``        ``}``        ``else` `{` `            ``// Increment evenCnt by 1``            ``oddCnt++;``        ``}``    ``}` `    ``// Print the minimum of oddCnt``    ``// and evenCnt``    ``cout << min(oddCnt, evenCnt);``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 4, 1, 2 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``minimumOperations(arr, N);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``public` `class` `GFG``{` `  ``// Function to find the minimum increments``  ``// required to difference between all``  ``// pairs of array elements even``  ``static` `void` `minimumOperations(``int``[] arr, ``int` `N)``  ``{` `    ``// Store the count of odd``    ``// and even numbers``    ``int` `oddCnt = ``0``, evenCnt = ``0``;` `    ``// Traverse the array``    ``for` `(``int` `i = ``0``; i < N; i++)``    ``{``      ``if` `(arr[i] % ``2` `== ``0``)``      ``{` `        ``// Increment evenCnt by 1``        ``evenCnt++;``      ``}``      ``else``      ``{` `        ``// Increment oddCnt by 1``        ``oddCnt++;``      ``}``    ``}` `    ``// Print the minimum of oddCnt``    ``// and evenCnt``    ``System.out.print(Math.min(oddCnt, evenCnt));``  ``}  ` `  ``// Driver code``  ``public` `static` `void` `main(String args[])``  ``{``    ``int``[] arr = { ``4``, ``1``, ``2` `};``    ``int` `N = arr.length;``    ``minimumOperations(arr, N);``  ``}``}` `// This code is contributed by AnkThon`

## Python3

 `# Python program for the above approach` `# Function to find the minimum increments``# required to difference between all``# pairs of array elements even``def` `minimumOperations(arr, N) :``    ` `    ``# Store the count of odd``    ``# and even numbers``    ``oddCnt ``=` `0``    ``evenCnt ``=` `0` `    ``# Traverse the array``    ``for` `i ``in` `range``(N):``        ``if` `(arr[i] ``%` `2` `=``=` `0``) :` `            ``# Increment evenCnt by 1``            ``evenCnt ``+``=` `1``        ` `        ``else` `:` `            ``# Increment evenCnt by 1``            ``oddCnt ``+``=` `1``        ` `    ``# Print minimum of oddCnt``    ``# and evenCnt``    ``print``(``min``(oddCnt, evenCnt))` `# Driver Code``arr ``=` `[ ``4``, ``1``, ``2` `]``N ``=` `len``(arr)``minimumOperations(arr, N)` `# This code is contributed by sanjoy_62.`

## C#

 `// C# program for the above approach``using` `System;``using` `System.Collections.Generic;``class` `GFG``{``    ` `    ``// Function to find the minimum increments``    ``// required to difference between all``    ``// pairs of array elements even``    ``static` `void` `minimumOperations(``int``[] arr, ``int` `N)``    ``{``      ` `        ``// Store the count of odd``        ``// and even numbers``        ``int` `oddCnt = 0, evenCnt = 0;``      ` `        ``// Traverse the array``        ``for` `(``int` `i = 0; i < N; i++)``        ``{``            ``if` `(arr[i] % 2 == 0)``            ``{``      ` `                ``// Increment evenCnt by 1``                ``evenCnt++;``            ``}``            ``else``            ``{``      ` `                ``// Increment evenCnt by 1``                ``oddCnt++;``            ``}``        ``}``      ` `        ``// Print the minimum of oddCnt``        ``// and evenCnt``        ``Console.Write(Math.Min(oddCnt, evenCnt));``    ``}  ` `  ``// Driver code``  ``static` `void` `Main()``  ``{``    ``int``[] arr = { 4, 1, 2 };``    ``int` `N = arr.Length;``    ``minimumOperations(arr, N);``  ``}``}` `// This code is contributed by divyeshrabadiya07.`

## Javascript

 ``

Output:

`1`

Time Complexity: O(N)
Auxiliary Space: O(1)

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