# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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 `

Output:

```2
```

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