# Maximum XOR of Two Numbers in an Array

Given an array Arr of non-negative integers of size N. The task is to find the maximum possible xor between two numbers present in the array.

Example

Input: Arr = {25, 10, 2, 8, 5, 3}
Output: 28
The maximum result is 5 ^ 25 = 28

Input: Arr = {1, 2, 3, 4, 5, 6, 7}
Output:
The maximum result is 1 ^ 6 = 7

Naive Approach: A Simple Solution is to generate all pairs of the given array and compute XOR of the pairs. Finally, return maximum XOR value. This solution takes

O(N^{2})

time.

Below is the implementation of the above approach:

C++

``````// C++ implementation
#include <bits/stdc++.h>
using namespace std;

// Function to return the
// maximum xor
int max_xor(int arr[], int n)
{

int maxXor = 0;

// Calculating xor of each pair
for (int i = 0; i < n; i++)
{
for (int j = i + 1; j < n; j++)
{
maxXor = max(maxXor, arr[i] ^ arr[j]);
}
}
return maxXor;
}

// Driver Code
int main()
{
int arr[] = { 25, 10, 2, 8, 5, 3 };
int n = sizeof(arr) / sizeof(arr);

// Function call
cout << max_xor(arr, n) << endl;
return 0;
}``````

Java

``````// Java implementation of the approach
class GFG
{

// Function to return the
// maximum xor
static int max_xor(int arr[], int n)
{
int maxXor = 0;

// Calculating xor of each pair
for (int i = 0; i < n; i++)
{
for (int j = i + 1; j < n; j++)
{
maxXor = Math.max(maxXor, arr[i] ^ arr[j]);
}
}
return maxXor;
}

// Driver Code
public static void main(String[] args)
{
int arr[] = { 25, 10, 2, 8, 5, 3 };
int n = arr.length;

// Function call
System.out.println(max_xor(arr, n));
}
}

// This code is contributed by Rajput-Ji``````

Python3

``````# Python3 implementation

# Function to return the
# maximum xor

def max_xor(arr, n):

maxXor = 0

# Calculating xor of each pair
for i in range(n):
for j in range(i + 1, n):
maxXor = max(maxXor,
arr[i] ^ arr[j])

return maxXor

# Driver Code
if __name__ == '__main__':

arr = [25, 10, 2, 8, 5, 3]
n = len(arr)

# Function call
print(max_xor(arr, n))

# This code is contributed by 29AjayKumar
``````

C#

``````// C# implementation of the approach
using System;

class GFG {

// Function to return the
// maximum xor
static int max_xor(int[] arr, int n)
{
int maxXor = 0;

// Calculating xor of each pair
for (int i = 0; i < n; i++)
{
for (int j = i + 1; j < n; j++)
{
maxXor = Math.Max(maxXor, arr[i] ^ arr[j]);
}
}
return maxXor;
}

// Driver Code
public static void Main()
{
int[] arr = { 25, 10, 2, 8, 5, 3 };
int n = arr.Length;

// Function call
Console.WriteLine(max_xor(arr, n));
}
}

// This code is contributed by AnkitRai01``````