Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

Remove one element to get maximum XOR

  • Difficulty Level : Medium
  • Last Updated : 03 Mar, 2022

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:
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:
 

 

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)


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!