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++
// 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 arrayint 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 Codeint 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 approachimport 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 arraydef 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 Codeif __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 approachusing 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
<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)
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