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

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: 1 
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 : 1 
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++ program for the above approach
#include <bits/stdc++.h>
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 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 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# 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

<script>

// JavaScript program for the above approach

    // Function to find the minimum
    // value of XOR of AND and OR of
    // any pair in the given array
    function maxAndXor(arr, n)
    {
        let ans = Number.MAX_VALUE;

        // Sort the array
        arr.sort();

        // Traverse the array arr[]
        for (let 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

    // Given array arr[]
        let arr = [ 1, 2, 3, 4, 5 ];

        let N = arr.length;

        // Function Call
        document.write(maxAndXor(arr, N));
 
 // This code is contributed by sanjoy_62.
</script>

1

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