Maximum XOR of Two Numbers in an Array | Set 2
Given an array arr[] consisting of N integers, the task is to find the maximum Bitwise XOR from all the possible pairs in the given array.
Examples:
Input: arr[] = {25, 10, 2, 8, 5, 3}
Output: 28
Explanation:
The maximum result is 5^25 = 28.Input: arr[] = {1, 2, 3, 4, 5, 6, 7}
Output: 7
Explanation:
The maximum result is 1^6 = 7.
Naive Approach: Refer to the article Maximum XOR of Two Numbers in an Array for the simplest approach to solve the problem by generating all pairs of the given array and computing XOR of each pair to find the maximum among them.
Time Complexity: O(N2)
Auxiliary Space: O(1)
Bitmasking Approach: Refer to the article Maximum XOR of Two Numbers in an Array to solve the problem using Bitmasking.
Time Complexity: O(N*log M), where M is the maximum number present in the array
Auxiliary Space: O(N)
Efficient Approach: The above approach can be solved by using Trie by inserting the binary representation of the numbers in the array arr[]. Now iterate the binary representation of all the elements in the array arr[] and if the current bit is 0 then find the path with value 1 or vice-versa in the Trie to get the maximum value of Bitwise XOR. Update the maximum value for each number. Below are the steps:
- Initialize maximumXOR as 0.
- Insert the binary representation of all the numbers in the given array arr[] in the Tree. While inserting in the trie if the current bit 0 then create a node in the left else create a node in the right of the current head node.
- Now, traverse the given array and for each element do the following:
- Initialize the currentXOR value as 0.
- Traverse the binary representation of the current number.
- If ith bit is 1 and node->left exists then update currentXOR as currentXOR + pow(2, i) and update node as node->left. Else update node = node->right.
- If ith bit is 0, and node->right exists then update currentXOR as currentXOR + pow(2, i) and update node as node->right. Else update node = node->left.
- For each array element in the above step, update the maximumXOR value if maximumXOR is greater than currentXOR.
- Print the value of maximumXOR after the above steps.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Structure of Trieclass node {public: node* left; node* right;};// Function to insert binary// representation of element x// in the Trievoid insert(int x, node* head){ // Store the head node* curr = head; for (int i = 30; i >= 0; i--) { // Find the i-th bit int val = (x >> i) & 1; if (val == 0) { // If curr->left is NULL if (!curr->left) curr->left = new node(); // Update curr to curr->left curr = curr->left; } else { // If curr->right is NULL if (!curr->right) curr->right = new node(); // Update curr to curr->right curr = curr->right; } }}// Function that finds the maximum// Bitwise XOR value for all such pairsint findMaximumXOR(int arr[], int n){ // head Node of Trie node* head = new node(); // Insert each element in trie for (int i = 0; i < n; i++) { insert(arr[i], head); } // Stores the maximum XOR value int ans = 0; // Traverse the given array for (int i = 0; i < n; i++) { // Stores the XOR with current // value arr[i] int curr_xor = 0; int M = pow(2, 30); node* curr = head; for (int j = 30; j >= 0; j--) { // Finding ith bit int val = (arr[i] >> j) & 1; // Check if the bit is 0 if (val == 0) { // If right node exists if (curr->right) { // Update the currentXOR curr_xor += M; curr = curr->right; } else { curr = curr->left; } } else { // Check if left node exists if (curr->left) { // Update the currentXOR curr_xor += M; curr = curr->left; } else { curr = curr->right; } } // Update M to M/2 for next set bit M /= 2; } // Update the maximum XOR ans = max(ans, curr_xor); } // Return the maximum XOR found return ans;}// Driver Codeint main(){ // Given array arr[] int arr[] = { 1, 2, 3, 4 }; int N = sizeof(arr) / sizeof(arr[0]); // Function Call cout << findMaximumXOR(arr, N); return 0;} |
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Java
// Java program for the above approachimport java.util.*;class GFG{// Structure of Triestatic class node{ node left; node right;};// Function to insert binary// representation of element x// in the Triestatic void insert(int x, node head){ // Store the head node curr = head; for(int i = 30; i >= 0; i--) { // Find the i-th bit int val = (x >> i) & 1; if (val == 0) { // If curr.left is null if (curr.left == null) curr.left = new node(); // Update curr to curr.left curr = curr.left; } else { // If curr.right is null if (curr.right == null) curr.right = new node(); // Update curr to curr.right curr = curr.right; } }}// Function that finds the maximum// Bitwise XOR value for all such pairsstatic int findMaximumXOR(int arr[], int n){ // head Node of Trie node head = new node(); // Insert each element in trie for(int i = 0; i < n; i++) { insert(arr[i], head); } // Stores the maximum XOR value int ans = 0; // Traverse the given array for(int i = 0; i < n; i++) { // Stores the XOR with current // value arr[i] int curr_xor = 0; int M = (int)Math.pow(2, 30); node curr = head; for(int j = 30; j >= 0; j--) { // Finding ith bit int val = (arr[i] >> j) & 1; // Check if the bit is 0 if (val == 0) { // If right node exists if (curr.right != null) { // Update the currentXOR curr_xor += M; curr = curr.right; } else { curr = curr.left; } } else { // Check if left node exists if (curr.left != null) { // Update the currentXOR curr_xor += M; curr = curr.left; } else { curr = curr.right; } } // Update M to M/2 for next set bit M /= 2; } // Update the maximum XOR ans = Math.max(ans, curr_xor); } // Return the maximum XOR found return ans;}// Driver Codepublic static void main(String[] args){ // Given array arr[] int arr[] = { 1, 2, 3, 4 }; int N = arr.length; // Function call System.out.print(findMaximumXOR(arr, N));}}// This code is contributed by Amit Katiyar |
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C#
// C# program for // the above approachusing System;class GFG{// Structure of Treepublic class node{ public node left; public node right;};// Function to insert binary// representation of element// x in the Treestatic void insert(int x, node head){ // Store the head node curr = head; for(int i = 30; i >= 0; i--) { // Find the i-th bit int val = (x >> i) & 1; if (val == 0) { // If curr.left is null if (curr.left == null) curr.left = new node(); // Update curr to curr.left curr = curr.left; } else { // If curr.right is null if (curr.right == null) curr.right = new node(); // Update curr to curr.right curr = curr.right; } }}// Function that finds the maximum// Bitwise XOR value for all // such pairsstatic int findMaximumXOR(int []arr, int n){ // Head Node of Tree node head = new node(); // Insert each element in tree for(int i = 0; i < n; i++) { insert(arr[i], head); } // Stores the maximum XOR value int ans = 0; // Traverse the given array for(int i = 0; i < n; i++) { // Stores the XOR with // current value arr[i] int curr_xor = 0; int M = (int)Math.Pow(2, 30); node curr = head; for(int j = 30; j >= 0; j--) { // Finding ith bit int val = (arr[i] >> j) & 1; // Check if the bit is 0 if (val == 0) { // If right node exists if (curr.right != null) { // Update the currentXOR curr_xor += M; curr = curr.right; } else { curr = curr.left; } } else { // Check if left node exists if (curr.left != null) { // Update the currentXOR curr_xor += M; curr = curr.left; } else { curr = curr.right; } } // Update M to M/2 // for next set bit M /= 2; } // Update the maximum XOR ans = Math.Max(ans, curr_xor); } // Return the maximum // XOR found return ans;}// Driver Codepublic static void Main(String[] args){ // Given array []arr int []arr = {1, 2, 3, 4}; int N = arr.Length; // Function call Console.Write(findMaximumXOR(arr, N));}}// This code is contributed by Rajput-Ji |
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Output:
7
Time Complexity: O(32*N)
Auxiliary Space: O(32*N)
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