# Minimum operations to make XOR of array zero

• Difficulty Level : Hard
• Last Updated : 19 Jul, 2022

We are given an array of n elements. The task is to make XOR of whole array 0. We can do the following to achieve this.

1. We can select any one of the elements.
2. After selecting an element, we can either increment or decrement it by 1.

We need to find the minimum number of increment/decrement operations required for the selected element to make the XOR sum of the whole array zero.

Examples:

```Input : arr[] = {2, 4, 8}
Output : Element = 8,
Operation required = 2
Explanation : Select 8 as element and perform 2
time decrement on it. So that it
became 6, Now our array is {2, 4, 6}
whose XOR sum is 0.

Input : arr[] = {1, 1, 1, 1}
Output : Element = 1,
Operation required = 0
Explanation : Select any of 1 and you have already
your XOR sum = 0. So, no operation
required.```

Naive Approach: Select an element and then find the XOR of the rest of the array. If that element became equals to XOR obtained then our XOR of the whole array should become zero. Now, our cost for that will be the absolute difference between the selected element and obtained XOR. This process of finding cost will be done for each element and thus resulting in Time Complexity of (n^2).

Efficient Approach: Find the XOR of the whole array. Now, suppose we have selected element arr[i], so cost required for that element will be absolute(arr[i]-(XORsum^arr[i])). Calculating the minimum of these absolute values for each element will be our minimum required operation also the element corresponding to the minimum required operation will be our selected element.

Implementation:

## C++

 `// CPP to find min cost to make``// XOR of whole array zero``#include ``using` `namespace` `std;` `// function to find min cost``void` `minCost(``int` `arr[], ``int` `n)``{``    ``int` `cost = INT_MAX;``    ``int` `element;` `    ``// calculate XOR sum of array``    ``int` `XOR = 0;``    ``for` `(``int` `i = 0; i < n; i++)``        ``XOR ^= arr[i];` `    ``// find the min cost and element corresponding``    ``for` `(``int` `i = 0; i < n; i++) {``        ``if` `(cost > ``abs``((XOR ^ arr[i]) - arr[i])) {``            ``cost = ``abs``((XOR ^ arr[i]) - arr[i]);``            ``element = arr[i];``        ``}``    ``}` `    ``cout << ``"Element = "` `<< element << endl;``    ``cout << ``"Operation required = "` `<< ``abs``(cost);``}` `// driver program``int` `main()``{``    ``int` `arr[] = { 2, 8, 4, 16 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);``    ``minCost(arr, n);``    ``return` `0;``}`

## Java

 `// JAVA program to find min cost to make``// XOR of whole array zero``import` `java.lang.*;` `class` `GFG``{``    ``// function to find min cost``    ``static` `void` `minCost(``int``[] arr, ``int` `n)``    ``{``        ``int` `cost = Integer.MAX_VALUE;``        ``int` `element=``0``;` `        ``// calculate XOR sum of array``        ``int` `XOR = ``0``;``        ``for` `(``int` `i = ``0``; i < n; i++)``            ``XOR ^= arr[i];` `        ``// find the min cost and element``        ``// corresponding``        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``if` `(cost > Math.abs((XOR ^ arr[i])``                                ``- arr[i])) {``                ``cost = Math.abs((XOR ^ arr[i]) -``                                       ``arr[i]);``                ``element = arr[i];``            ``}``        ``}` `    ``System.out.println(``"Element = "` `+ element);``    ``System.out.println(``"Operation required = "``+``                             ``Math.abs(cost));``    ``}` `    ``// driver program``    ``public` `static` `void` `main (String[] args)``    ``{``        ``int``[] arr = { ``2``, ``8``, ``4``, ``16` `};``        ``int` `n = arr.length;``        ``minCost(arr, n);``    ``}``}``/* This code is contributed by Kriti Shukla */`

## Python3

 `# python to find min cost to make``# XOR of whole array zero` `# function to find min cost``def` `minCost(arr,n):``    ` `    ``cost ``=` `999999``;``    ` `    ``# calculate XOR sum of array``    ``XOR ``=` `0``;``    ``for` `i ``in` `range``(``0``, n):``        ``XOR ^``=` `arr[i];` `    ``# find the min cost and element``    ``# corresponding``    ``for` `i ``in` `range``(``0``,n):``        ``if` `(cost > ``abs``((XOR ^ arr[i]) ``-` `arr[i])):``            ``cost ``=` `abs``((XOR ^ arr[i]) ``-` `arr[i])``            ``element ``=` `arr[i]` `    ``print``(``"Element = "``, element)``    ``print``(``"Operation required = "``, ``abs``(cost))`  `# driver program``arr ``=` `[ ``2``, ``8``, ``4``, ``16` `]``n ``=` `len``(arr)``minCost(arr, n)` `# This code is contributed by Sam007`

## C#

 `// C# program to find min cost to``// make XOR of whole array zero``using` `System;` `class` `GFG``{``    ``// function to find min cost``    ``static` `void` `minCost(``int` `[]arr, ``int` `n)``    ``{``        ``int` `cost = ``int``.MaxValue;``        ``int` `element=0;` `        ``// calculate XOR sum of array``        ``int` `XOR = 0;``        ``for` `(``int` `i = 0; i < n; i++)``            ``XOR ^= arr[i];` `        ``// find the min cost and``        ``// element corresponding``        ``for` `(``int` `i = 0; i < n; i++)``        ``{``            ``if` `(cost > Math.Abs((XOR ^ arr[i]) - arr[i]))``            ``{``                ``cost = Math.Abs((XOR ^ arr[i]) - arr[i]);``                ``element = arr[i];``            ``}``        ``}` `    ``Console.WriteLine(``"Element = "` `+ element);``    ``Console.Write(``"Operation required = "``+``                          ``Math.Abs(cost));``    ``}` `    ``// Driver program``    ``public` `static` `void` `Main ()``    ``{``        ``int` `[]arr = {2, 8, 4, 16};``        ``int` `n = arr.Length;``        ``minCost(arr, n);``    ``}``}` `// This code is contributed by nitin mittal.`

## PHP

 ` ``abs``((``\$XOR` `^ ``\$arr``[``\$i``]) -``                                 ``\$arr``[``\$i``]))``        ``{``            ``\$cost` `= ``abs``((``\$XOR` `^ ``\$arr``[``\$i``]) -``                                 ``\$arr``[``\$i``]);``            ``\$element` `= ``\$arr``[``\$i``];``        ``}``    ``}` `    ``echo` `"Element = "` `, ``\$element` `,``"\n"``;``    ``echo` `"Operation required = "` `, ``abs``(``\$cost``);``}` `// Driver Code``\$arr` `= ``array``(2, 8, 4, 16) ;``\$n` `= ``count``(``\$arr``);``minCost(``\$arr``, ``\$n``);` `// This code is contributed by vt_m.``?>`

## Javascript

 ``

Output

```Element = 16
Operation required = 2```

Time Complexity : O(n)