Remove one element to get maximum XOR

Given an array arr[] of N elements, the task is to remove one element from the array such that the XOR value of the array is maximized. Print the maximized value.
Examples:

Input: arr[] = {1, 1, 3} 
Output: 2 
All possible ways of deleting one element and their corresponding XOR values will be: 
a) Remove 1 -> (1 XOR 3) = 2 
b) Remove 1 -> (1 XOR 3) = 2 
c) Remove 3 -> (1 XOR 1) = 0 
Thus, the final answer is 2.
Input: arr[] = {3, 3, 3} 
Output: 0 

Naive approach: One way will be to remove each element one by one and then finding the XOR of the remaining elements. The time complexity of this approach will be O(N²).
Efficient approach:

• Find XOR of all the elements of the array. Let’s call this value X.
• For each element arr[i], perform Y = (X XOR arr[i]) and update the final answer as ans = max(Y, ans).

The above method works because if (A XOR B) = C then (C XOR B) = A. To find XOR(arr[0…i-1]) ^ XOR(arr[i+1…N-1]), all we have to perform is XOR(arr) ^ arr[i] where XOR(arr) is the XOR of all the elements of the array.
Below is the implementation of the above approach: 

C++

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

// Function to return the maximized XOR
// after removing an element from the array
int maxXOR(int* arr, int n)
{
    // Find XOR of the complete array
    int xorArr = 0;
    for (int i = 0; i < n; i++)
        xorArr ^= arr[i];

    // To store the final answer
    int ans = 0;

    // Iterating through the array to find
    // the final answer
    for (int i = 0; i < n; i++)
        ans = max(ans, (xorArr ^ arr[i]));

    // Return the final answer
    return ans;
}

// Driver code
int main()
{
    int arr[] = { 1, 1, 3 };
    int n = sizeof(arr) / sizeof(int);

    cout << maxXOR(arr, n);

    return 0;
}

Java

// Java implementation of the approach
class GFG
{

    // Function to return the maximized XOR
    // after removing an element from the array
    static int maxXOR(int arr[], int n)
    {
        // Find XOR of the complete array
        int xorArr = 0;
        for (int i = 0; i < n; i++)
            xorArr ^= arr[i];

        // To store the final answer
        int ans = 0;

        // Iterating through the array to find
        // the final answer
        for (int i = 0; i < n; i++)
            ans = Math.max(ans, (xorArr ^ arr[i]));

        // Return the final answer
        return ans;
    }

    // Driver code
    public static void main (String[] args)
    {
        int arr[] = { 1, 1, 3 };
        int n = arr.length;
        System.out.println(maxXOR(arr, n));
    }
}

// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach

# Function to return the maximized XOR
# after removing an element from the array
def maxXOR(arr, n):

    # Find XOR of the complete array
    xorArr = 0
    for i in range(n):
        xorArr ^= arr[i]

    # To store the final answer
    ans = 0

    # Iterating through the array to find
    # the final answer
    for i in range(n):
        ans = max(ans, (xorArr ^ arr[i]))

    # Return the final answer
    return ans

# Driver code
arr = [1, 1, 3]
n = len(arr)

print(maxXOR(arr, n))

# This code is contributed by Mohit Kumar

C#

// C# implementation of the approach
using System;

class GFG
{

    // Function to return the maximized XOR
    // after removing an element from the array
    static int maxXOR(int []arr, int n)
    {
        // Find XOR of the complete array
        int xorArr = 0;
        for (int i = 0; i < n; i++)
            xorArr ^= arr[i];

        // To store the readonly answer
        int ans = 0;

        // Iterating through the array to find
        // the readonly answer
        for (int i = 0; i < n; i++)
            ans = Math.Max(ans, (xorArr ^ arr[i]));

        // Return the readonly answer
        return ans;
    }

    // Driver code
    public static void Main(String[] args)
    {
        int []arr = { 1, 1, 3 };
        int n = arr.Length;
        Console.WriteLine(maxXOR(arr, n));
    }
}

// This code is contributed by 29AjayKumar

Javascript

<script>
// Javascript implementation of the approach

// Function to return the maximized XOR
// after removing an element from the array
function maxXOR(arr, n)
{
    // Find XOR of the complete array
    let xorArr = 0;
    for (let i = 0; i < n; i++)
        xorArr ^= arr[i];

    // To store the final answer
    let ans = 0;

    // Iterating through the array to find
    // the final answer
    for (let i = 0; i < n; i++)
        ans = Math.max(ans, (xorArr ^ arr[i]));

    // Return the final answer
    return ans;
}

// Driver code
    let arr = [ 1, 1, 3 ];
    let n = arr.length;

    document.write(maxXOR(arr, n));

</script>

Output:

2

Time Complexity: O(n)

Auxiliary Space: O(1)