# Minimum XOR of OR and AND of any pair in the Array

• Last Updated : 05 May, 2021

Given an array arr[] of N positive integers the task is to find the minimum value of Bitwise XOR of Bitwise OR and AND of any pair in the given array.

Examples:

Input: arr[] = {1, 2, 3, 4, 5}
Output:
Explanation:
For element 2 & 3:
The value of the expression (2&3) xor (2|3) is 1, which is the minimum from all the pairs in the given array.
Input : arr[] = {3, 6, 8, 4, 5}
Output :
Explanation:
For element 4 & 5:
The value of the expression (4&5) xor (4|5) is 1, which is the minimum from all the pairs in the given array.

Approach: For any pairs of elements say X and Y, the value of the expression (X&Y) xor (X|Y) can be written as:

=> (X.Y)^(X+Y)
=> (X.Y)(X+Y)’ + (X.Y)'(X+Y)
=> (X.Y)(X’.Y’) + (X’+Y’)(X+Y)
=> X.X’.Y.Y’ + X’.X + X’.Y + Y’.X + Y’.Y
=> 0 + 0 + X’.Y + Y’.X + 0
=> X^Y

From the above calculations, for any pairs (X, Y) in the given array, the expression (X&Y) xor (X|Y) is reduced to X xor Y. Therefore, the minimum value of Bitwise XOR of Bitwise OR and AND of any pair in the given array is given XOR of minimum value pair which can be calculated using the approach discussed in this article.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to find the minimum``// value of XOR of AND and OR of``// any pair in the given array``int` `maxAndXor(``int` `arr[], ``int` `n)``{``    ``int` `ans = INT_MAX;` `    ``// Sort the array``    ``sort(arr, arr + n);` `    ``// Traverse the array arr[]``    ``for` `(``int` `i = 0; i < n - 1; i++) {` `        ``// Compare and Find the minimum``        ``// XOR value of an array.``        ``ans = min(ans, arr[i] ^ arr[i + 1]);``    ``}` `    ``// Return the final answer``    ``return` `ans;``}` `// Driver Code``int` `main()``{``    ``// Given array``    ``int` `arr[] = { 1, 2, 3, 4, 5 };` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    ``// Function Call``    ``cout << maxAndXor(arr, N);``    ``return` `0;``}`

## Java

 `// Java program to for above approach``import` `java.io.*;``import` `java.util.Arrays;``class` `GFG {` `    ``// Function to find the minimum``    ``// value of XOR of AND and OR of``    ``// any pair in the given array``    ``static` `int` `maxAndXor(``int` `arr[], ``int` `n)``    ``{``        ``int` `ans = Integer.MAX_VALUE;` `        ``// Sort the array``        ``Arrays.sort(arr);` `        ``// Traverse the array arr[]``        ``for` `(``int` `i = ``0``; i < n - ``1``; i++) {` `            ``// Compare and Find the minimum``            ``// XOR value of an array.``            ``ans = Math.min(ans, arr[i] ^ arr[i + ``1``]);``        ``}` `        ``// Return the final answer``        ``return` `ans;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``// Given array arr[]``        ``int` `arr[] = ``new` `int``[] { ``1``, ``2``, ``3``, ``4``, ``5` `};` `        ``int` `N = arr.length;` `        ``// Function Call``        ``System.out.println(maxAndXor(arr, N));``    ``}``}` `// This code is contributed by Shubham Prakash`

## Python3

 `# Python3 program for the above approach` `# Function to find the minimum``# value of XOR of AND and OR of``# any pair in the given array`  `def` `maxAndXor(arr, n):` `    ``ans ``=` `float``(``'inf'``)` `    ``# Sort the array``    ``arr.sort()` `    ``# Traverse the array arr[]``    ``for` `i ``in` `range``(n ``-` `1``):` `        ``# Compare and Find the minimum``        ``# XOR value of an array.``        ``ans ``=` `min``(ans, arr[i] ^ arr[i ``+` `1``])` `    ``# Return the final answer``    ``return` `ans`  `# Driver Code``if` `__name__ ``=``=` `'__main__'``:` `    ``# Given array``    ``arr ``=` `[``1``, ``2``, ``3``, ``4``, ``5``]``    ``N ``=` `len``(arr)` `    ``# Function Call``    ``print``(maxAndXor(arr, N))` `# This code is contributed by Shivam Singh`

## C#

 `// C# program to for above approach``using` `System;``class` `GFG {` `    ``// Function to find the minimum``    ``// value of XOR of AND and OR of``    ``// any pair in the given array``    ``static` `int` `maxAndXor(``int``[] arr, ``int` `n)``    ``{``        ``int` `ans = 9999999;` `        ``// Sort the array``        ``Array.Sort(arr);` `        ``// Traverse the array arr[]``        ``for` `(``int` `i = 0; i < n - 1; i++) {` `            ``// Compare and Find the minimum``            ``// XOR value of an array.``            ``ans = Math.Min(ans, arr[i] ^ arr[i + 1]);``        ``}` `        ``// Return the final answer``        ``return` `ans;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main()``    ``{``        ``// Given array arr[]``        ``int``[] arr = ``new` `int``[] { 1, 2, 3, 4, 5 };` `        ``int` `N = arr.Length;` `        ``// Function Call``        ``Console.Write(maxAndXor(arr, N));``    ``}``}` `// This code is contributed by Code_Mech`

## Javascript

 ``
Output:
`1`

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

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