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 codeint 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 approachclass 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 codeA = [ 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 approachusing 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> |
100
Time Complexity: O(N)
Space Complexity: O(1)
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