在第二个数组中查找额外的元素

给定两个数组A []和B [] 。第二个数组B []包含A []的所有元素，除了1个额外的元素。任务是找到那个额外的元素。
例子：

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 
 

简单方法：运行嵌套循环，并找到B中的元素[]，这是不存在于A []。这种方法的时间复杂度将是O(n ² ) 。
有效的方法：如果A []和B []进行异或在一起，然后A []中的每个元素将给予0与其在B []发生和额外的元件比如X当异或与0会给的所有元素(X XOR 0)= X ，这就是结果。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
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

// 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.

输出：

100

时间复杂度： O(N)
空间复杂度： O(1)

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