# Minimum absolute difference of XOR values of two subarrays

• Difficulty Level : Medium
• Last Updated : 30 Apr, 2021

Given an array containing n numbers. The problem is to split the array into two subarrays such that the absolute difference of the xor values of the two subarrays is minimum.
Note: Array contains at least 2 numbers.

Examples:

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```Input : arr[] = {12, 6, 20, 14, 38, 6}
Output : 16
The two subarrays are:
{12, 6, 20} = 12 ^ 6 ^ 20 = 30
{14, 38, 6} = 14 ^ 38 ^ 6 = 46
Absolute difference = abs(30-46)
= 16

Input : arr[] = {10, 16, 9, 34, 7, 46, 23}
Output : 1```

Naive Approach: Using two for loops find the xor values of the two subarrays sub_arr1[1..i] and sub_arr2[i+1..n] for every i = 1 to n-1. Find their absolute difference and accordingly update the minimum absolute difference.
Time Complexity: O(n2)

Efficient Approach: It is based on the following properties of the ^(xor) operator:

1. a ^ 0 = a.
2. a ^ a = 0.

Algorithm:

```minDiffBtwXorValues(arr, n)
Declare tot_xor = 0
for i = 0 to n-1
tot_xor ^= arr[i]

Declare part_xor = 0
Declare min = Maximum Integer

for i = 0 to n-2
tot_xor ^= arr[i]
part_xor ^= arr[i]
if abs(tot_xor - part_xor) < min
min = abs(tot_xor - part_xor)

return min```

## C++

 `// C++ implementation to find the minimum``// absolute difference of the xor values``// of the two subarrays``#include ``using` `namespace` `std;` `// function to find the minimum absolute``// difference of the xor values of the``// two subarrays``int` `minDiffBtwXorValues(``int` `arr[], ``int` `n)``{``    ``// to store the xor value of the``    ``// entire array``    ``int` `tot_xor = 0;``    ``for` `(``int` `i = 0; i < n; i++)``        ``tot_xor ^= arr[i];` `    ``// 'part_xor' to store the xor value``    ``// of some subarray``    ``int` `part_xor = 0, min = INT_MAX;` `    ``for` `(``int` `i = 0; i < n - 1; i++) {` `        ``// removing the xor value of the``        ``// subarray [0..i] form 'tot_xor',``        ``// i.e, it will contain the xor``        ``// value of the subarray [i+1..n-1]``        ``tot_xor ^= arr[i];` `        ``// calculating the xor value of the``        ``// subarray [0..i]``        ``part_xor ^= arr[i];` `        ``// if absolute difference is minimum,``        ``// then update 'min'``        ``if` `(``abs``(tot_xor - part_xor) < min)``            ``min = ``abs``(tot_xor - part_xor);``    ``}` `    ``// required minimum absolute difference``    ``return` `min;``}` `// Driver program to test above``int` `main()``{``    ``int` `arr[] = { 12, 6, 20, 14, 38, 6 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``cout << ``"Minimum Absolute Difference = "``         ``<< minDiffBtwXorValues(arr, n);``    ``return` `0;``}`

## Java

 `// Java implementation to``// find the minimum``// absolute difference``// of the xor values``// of the two subarrays` `import` `java.util.*;``import` `java.lang.*;` `public` `class` `GfG{` `    ``// function to find``    ``// the minimum absolute``    ``// difference of the``    ``// xor values of the``    ``// two subarrays``    ``public` `static` `int` `minDiffBtwXorValues(``int` `arr[],``    ``int` `n)``    ``{``        ``// to store the xor value of the``        ``// entire array``        ``int` `tot_xor = ``0``;``        ``for` `(``int` `i = ``0``; i < n; i++)``            ``tot_xor ^= arr[i];`` ` `        ``// 'part_xor' to store the xor value``        ``// of some subarray``        ``int` `part_xor = ``0``, min = Integer.MAX_VALUE;`` ` `        ``for` `(``int` `i = ``0``; i < n - ``1``; i++) {`` ` `            ``// removing the xor value of the``            ``// subarray [0..i] form 'tot_xor',``            ``// i.e, it will contain the xor``            ``// value of the subarray [i+1..n-1]``            ``tot_xor ^= arr[i];`` ` `            ``// calculating the xor value of the``            ``// subarray [0..i]``            ``part_xor ^= arr[i];`` ` `            ``// if absolute difference is minimum,``            ``// then update 'min'``            ``if` `(Math.abs(tot_xor - part_xor) < min)``                ``min = Math.abs(tot_xor - part_xor);``        ``}`` ` `        ``// required minimum absolute difference``        ``return` `min;``    ``}``    ` `    ``// Driver function``    ``public` `static` `void` `main(String argc[]){` `        ``int` `arr[] = { ``12``, ``6``, ``20``, ``14``, ``38``, ``6` `};``        ``int` `n = ``6``;``        ``System.out.println(``"Minimum Absolute Difference = "` `+``        ``minDiffBtwXorValues(arr, n));``    ``}``    ` `}` `// This code is contributed by Sagar Shukla`

## Python

 `# Python implementation to find the minimum``# absolute difference of the xor values``# of the two subarrays` `import` `sys` `# function to find the minimum absolute``# difference of the xor values of the``# two subarrays``def` `minDiffBtwXorValues(arr, n):``   ` `    ``# to store the xor value of the``    ``# entire array``    ``tot_xor ``=` `0``    ` `    ``for` `i ``in` `range``(n):``        ``tot_xor ^``=` `arr[i]`` ` `    ``# 'part_xor' to store the xor value``    ``# of some subarray``    ``part_xor ``=` `0``    ``min` `=` `sys.maxint`` ` `    ``for` `i ``in` `range``(n ``-` `1``):`` ` `        ``# removing the xor value of the``        ``# subarray [0..i] form 'tot_xor',``        ``# i.e, it will contain the xor``        ``# value of the subarray [i+1..n-1]``        ``tot_xor ^``=` `arr[i]`` ` `        ``# calculating the xor value of the``        ``# subarray [0..i]``        ``part_xor ^``=` `arr[i]`` ` `        ``# if absolute difference is minimum,``        ``# then update 'min'``        ``if` `(``abs``(tot_xor ``-` `part_xor) < ``min``):``            ``min` `=` `abs``(tot_xor ``-` `part_xor)`` ` `    ``# required minimum absolute difference``    ``return` `min`` ` `# Driver program to test above``arr ``=` `[ ``12``, ``6``, ``20``, ``14``, ``38``, ``6` `]``n ``=` `len``(arr)``print` `"Minimum Absolute Difference ="``, minDiffBtwXorValues(arr, n)` `# This code is contributed by Sachin Bisht`

## C#

 `// C# implementation to``// find the minimum``// absolute difference``// of the xor values``// of the two subarrays``using` `System;` `class` `GFG {``    ` `    ``// function to find``    ``// the minimum absolute``    ``// difference of the``    ``// xor values of the``    ``// two subarrays``    ``public` `static` `int` `minDiffBtwXorValues(``int``[] arr,``                                              ``int` `n)``    ``{``        ``// to store the xor value of the``        ``// entire array``        ``int` `tot_xor = 0;``        ``for` `(``int` `i = 0; i < n; i++)``            ``tot_xor ^= arr[i];` `        ``// 'part_xor' to store the xor value``        ``// of some subarray``        ``int` `part_xor = 0, min = ``int``.MaxValue;` `        ``for` `(``int` `i = 0; i < n - 1; i++) {` `            ``// removing the xor value of the``            ``// subarray [0..i] form 'tot_xor',``            ``// i.e, it will contain the xor``            ``// value of the subarray [i+1..n-1]``            ``tot_xor ^= arr[i];` `            ``// calculating the xor value of the``            ``// subarray [0..i]``            ``part_xor ^= arr[i];` `            ``// if absolute difference is minimum,``            ``// then update 'min'``            ``if` `(Math.Abs(tot_xor - part_xor) < min)``                ``min = Math.Abs(tot_xor - part_xor);``        ``}` `        ``// required minimum absolute difference``        ``return` `min;``    ``}` `    ``// Driver function``    ``public` `static` `void` `Main()``    ``{` `        ``int``[] arr = { 12, 6, 20, 14, 38, 6 };``        ``int` `n = 6;``        ``Console.WriteLine(``"Minimum Absolute Difference = "` `+``                               ``minDiffBtwXorValues(arr, n));``    ``}``}` `// This code is contributed by Sam007`

## PHP

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## Javascript

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Output:

`Minimum Absolute Difference = 16`

Time Complexity: O(n)

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