Find extra element in the second array

Given two arrays A[] and B[]. The second array B[] contains all the elements of A[] except for 1 extra element. The task is to find that extra element.
Examples:

Input: A[] = { 1, 2, 3 }, B[] = {1, 2, 3, 4} 
Output: 4 
Element 4 is not present in array
Input: A[] = {10, 15, 5}, B[] = {10, 100, 15, 5} 
Output: 100 

Naive approach: Run nested loops and find the element in B[] which is not present in A[]. The time complexity of this approach will be O(n²).
Efficient approach: If all the elements of the A[] and B[] are XORed together then every element of A[] will give 0 with its occurrence in B[] and the extra element say X when XORed with 0 will give (X XOR 0) = X which is the result.
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 extra
// element in B[]
int extraElement(int A[], int B[], int n)
{

    // To store the result
    int ans = 0;

    // Find the XOR of all the element
    // of array A[] and array B[]
    for (int i = 0; i < n; i++)
        ans ^= A[i];
    for (int i = 0; i < n + 1; i++)
        ans ^= B[i];

    return ans;
}

// Driver code
int main()
{
    int A[] = { 10, 15, 5 };
    int B[] = { 10, 100, 15, 5 };
    int n = sizeof(A) / sizeof(int);

    cout << extraElement(A, B, n);

    return 0;
}

Java

// Java implementation of the approach
class GFG
{

    // Function to return the extra
    // element in B[]
    static int extraElement(int A[], int B[], int n)
    {

        // To store the result
        int ans = 0;

        // Find the XOR of all the element
        // of array A[] and array B[]
        for (int i = 0; i < n; i++)
            ans ^= A[i];
        for (int i = 0; i < n + 1; i++)
            ans ^= B[i];

        return ans;
    }

    // Driver code
    public static void main (String[] args)
    {
        int A[] = { 10, 15, 5 };
        int B[] = { 10, 100, 15, 5 };
        int n = A.length;

        System.out.println(extraElement(A, B, n));
    }
}

// This code is contributed by kanugargng

Python3

# Python3 implementation of the approach
# Function to return the extra
# element in B[]
def extraElement(A, B, n):

    # To store the result
    ans = 0;

    # Find the XOR of all the element
    # of array A[] and array B[]
    for i in range(n):
        ans ^= A[i];
    for i in range(n + 1):
        ans ^= B[i];

    return ans;

# Driver code
A = [ 10, 15, 5 ];
B = [ 10, 100, 15, 5 ];
n = len(A);

print(extraElement(A, B, n));

# This code is contributed by 29AjayKumar

C#

// C# implementation of the approach
using System;

class GFG
{

    // Function to return the extra
    // element in B[]
    static int extraElement(int []A,
                            int []B, int n)
    {

        // To store the result
        int ans = 0;

        // Find the XOR of all the element
        // of array A[] and array B[]
        for (int i = 0; i < n; i++)
            ans ^= A[i];
        for (int i = 0; i < n + 1; i++)
            ans ^= B[i];

        return ans;
    }

    // Driver code
    public static void Main (String[] args)
    {
        int []A = { 10, 15, 5 };
        int []B = { 10, 100, 15, 5 };
        int n = A.Length;

        Console.WriteLine(extraElement(A, B, n));
    }
}

// This code is contributed by 29AjayKumar

Javascript

<script>
// Javascript implementation of the approach

// Function to return the extra
// element in B[]
function extraElement(A, B, n)
{

    // To store the result
    let ans = 0;

    // Find the XOR of all the element
    // of array A[] and array B[]
    for (let i = 0; i < n; i++)
        ans ^= A[i];
    for (let i = 0; i < n + 1; i++)
        ans ^= B[i];

    return ans;
}

// Driver code
    let A = [ 10, 15, 5 ];
    let B = [ 10, 100, 15, 5 ];
    let n = A.length;

    document.write(extraElement(A, B, n));

// This code is contributed by subhammahato348.
</script>

Output:

100

Time Complexity: O(N) 
Space Complexity: O(1)