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 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 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 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)
Approach 2: Using binary search
- Initialize two pointers low and high to the first and last index of the second array respectively.
- While low is less than or equal to high, repeat steps 3-5:
- Find the middle index mid of the search space by taking the average of low and high.
- Compare the element at index mid of the second array with the element at the same index in the first array.
- If they are equal, then the extra element is on the right side of the middle index. So, set low to mid + 1.
- If the elements at index mid are not equal, then the extra element is on the left side of the middle index. So, set high to mid-1.
- After the loop, return the element at the low index of the second array, as this is the extra element.
C++
#include <iostream>using namespace std;
int binarySearch(int arr[], int low, int high, int x) {
if (high >= low) {
int mid = low + (high - low) / 2;
if (arr[mid] == x) {
return mid;
}
if (arr[mid] > x) {
return binarySearch(arr, low, mid - 1, x);
}
return binarySearch(arr, mid + 1, high, x);
}
return -1;
}int findExtra(int arr1[], int arr2[], int n) {
int low = 0, high = n - 2;
while (low <= high) {
int mid = (low + high) / 2;
if (arr2[mid] == arr1[mid]) {
low = mid + 1;
} else {
high = mid - 1;
}
}
return arr2[low];
}int main() {
int arr1[] = { 10, 15, 5 };
int arr2[] = { 10, 100,15, 5 };
int n = sizeof(arr2) / sizeof(arr2[0]);
cout << findExtra(arr1, arr2, n) << endl;
return 0;
} |
Java
public class FindExtraElement {
public static int binarySearch(int[] arr, int low, int high, int x) {
if (high >= low) {
int mid = low + (high - low) / 2;
if (arr[mid] == x) {
return mid;
}
if (arr[mid] > x) {
return binarySearch(arr, low, mid - 1, x);
}
return binarySearch(arr, mid + 1, high, x);
}
return -1;
}
public static int findExtra(int[] arr1, int[] arr2, int n) {
int low = 0, high = n - 2;
while (low <= high) {
int mid = (low + high) / 2;
if (arr2[mid] == arr1[mid]) {
low = mid + 1;
} else {
high = mid - 1;
}
}
return arr2[low];
}
public static void main(String[] args) {
int[] arr1 = { 10, 15, 5 };
int[] arr2 = { 10, 100, 15, 5 };
int n = arr2.length;
System.out.println(findExtra(arr1, arr2, n));
}
} |
Python3
def binarySearch(arr, low, high, x):
if high >= low:
mid = low + (high - low) // 2
if arr[mid] == x:
return mid
if arr[mid] > x:
return binarySearch(arr, low, mid - 1, x)
return binarySearch(arr, mid + 1, high, x)
return -1
def findExtra(arr1, arr2, n):
low, high = 0, n - 2
while low <= high:
mid = (low + high) // 2
if arr2[mid] == arr1[mid]:
low = mid + 1
else:
high = mid - 1
return arr2[low]
arr1 = [10, 15, 5]
arr2 = [10, 100, 15, 5]
n = len(arr2)
print(findExtra(arr1, arr2, n))
|
C#
using System;
class Program
{ static int BinarySearch(int[] arr, int low, int high, int x)
{
if (high >= low)
{
int mid = low + (high - low) / 2;
if (arr[mid] == x)
{
return mid;
}
if (arr[mid] > x)
{
return BinarySearch(arr, low, mid - 1, x);
}
return BinarySearch(arr, mid + 1, high, x);
}
return -1;
}
static int FindExtra(int[] arr1, int[] arr2, int n)
{
int low = 0, high = n - 2;
while (low <= high)
{
int mid = (low + high) / 2;
if (arr2[mid] == arr1[mid])
{
low = mid + 1;
}
else
{
high = mid - 1;
}
}
return arr2[low];
}
static void Main()
{
int[] arr1 = { 10, 15, 5 };
int[] arr2 = { 10, 100, 15, 5 };
int n = arr2.Length;
Console.WriteLine(FindExtra(arr1, arr2, n));
}
} |
Javascript
function binarySearch(arr, low, high, x) {
if (high >= low) {
let mid = low + Math.floor((high - low) / 2);
if (arr[mid] === x) {
return mid;
}
if (arr[mid] > x) {
return binarySearch(arr, low, mid - 1, x);
}
return binarySearch(arr, mid + 1, high, x);
}
return -1;
}function findExtra(arr1, arr2, n) {
let low = 0, high = n - 2;
while (low <= high) {
let mid = Math.floor((low + high) / 2);
if (arr2[mid] === arr1[mid]) {
low = mid + 1;
} else {
high = mid - 1;
}
}
return arr2[low];
}let arr1 = [ 10, 15, 5 ];let arr2 = [ 10, 100, 15, 5 ];let n = arr2.length;console.log(findExtra(arr1, arr2, n)); |
Output
100
Time Complexity: O(logN)
Auxiliary Space: O(1)
Approach 3: Function to Find the Extra Element in Two Arrays by Calculating the Sum Difference.
- Calculate the sum of all elements in array A.
- Calculate the sum of all elements in array B.
- Return the difference between sum_A and sum_B as the missing element.
Here is the implementation of above approach:-
C++
#include <iostream>#include <vector>using namespace std;
int extraElement(vector<int> A, vector<int> B) {
int sum_A = 0, sum_B = 0;
for (int i = 0; i < A.size(); i++) {
sum_A += A[i];
}
for (int i = 0; i < B.size(); i++) {
sum_B += B[i];
}
return sum_B - sum_A;
}int main() {
vector<int> A = {10, 15, 5};
vector<int> B = {10, 100, 15, 5};
cout << extraElement(A, B) << endl;
return 0;
} |
Java
import java.util.*;
public class Main {
public static int extraElement(int[] A, int[] B) {
int sum_A = 0, sum_B = 0;
for (int i = 0; i < A.length; i++) {
sum_A += A[i];
}
for (int i = 0; i < B.length; i++) {
sum_B += B[i];
}
return sum_B - sum_A;
}
public static void main(String[] args) {
int[] A = {10, 15, 5};
int[] B = {10, 100, 15, 5};
System.out.println(extraElement(A, B));
}
} |
Python3
def extraElement(A, B):
sum_A = sum(A)
sum_B = sum(B)
return sum_B - sum_A
A = [10, 15, 5]
B = [10, 100, 15, 5]
print(extraElement(A, B))
|
C#
using System;
public class Program
{ public static int extraElement(int[] A, int[] B)
{
int sum_A = 0, sum_B = 0;
for (int i = 0; i < A.Length; i++)
{
sum_A += A[i];
}
for (int i = 0; i < B.Length; i++)
{
sum_B += B[i];
}
return sum_B - sum_A;
}
public static void Main()
{
int[] A = {10, 15, 5};
int[] B = {10, 100, 15, 5};
int n = A.Length;
Console.WriteLine(extraElement(A, B));
}
} |
Javascript
function extraElement(A, B) {
let sum_A = 0, sum_B = 0;
for (let i = 0; i < A.length; i++) {
sum_A += A[i];
}
for (let i = 0; i < B.length; i++) {
sum_B += B[i];
}
return sum_B - sum_A;
}let A = [10, 15, 5];let B = [10, 100, 15, 5];let n = A.length;console.log(extraElement(A, B)); |
Output
100
The time complexity of the function is O(N) because it needs to iterate over each element in both arrays to calculate their sums.
The space complexity of the function is O(1) because it only requires a constant amount of extra space to store the two sum variables.