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++

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// 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;
}

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Java














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// 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 



















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Python3














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# 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 



















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C#














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// 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 



















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Output:


100





Time Complexity: O(N)

Space Complexity: O(1)





    

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