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Check if the array has an element which is equal to XOR of remaining elements

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  • Last Updated : 05 Sep, 2022
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Given an array arr[] of N elements, the task is to check if the array has an element which is equal to the XOR of all the remaining elements.
Examples: 
 

Input: arr[] = { 8, 2, 4, 15, 1 } 
Output: Yes 
8 is the required element as 2 ^ 4 ^ 15 ^ 1 = 8.
Input: arr[] = {4, 2, 3} 
Output: No 
 

 

Approach: First, take the XOR of all the elements of the array and store it in a variable xorArr. Now traverse the complete array again and for every element, calculate the xor of the array elements excluding the current element arr[i] i.e. x = xorArr ^ arr[i]
If x = arr[i] then arr[i] is the required element and hence print Yes. If no such element is found then print No.
Below is the implementation of the above approach: 
 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function that returns true if the array
// contains an element which is equal to
// the XOR of the remaining elements
bool containsElement(int arr[], int n)
{
 
    // To store the XOR of all
    // the array elements
    int xorArr = 0;
    for (int i = 0; i < n; ++i)
        xorArr ^= arr[i];
 
    // For every element of the array
    for (int i = 0; i < n; ++i) {
 
        // Take the XOR after excluding
        // the current element
        int x = xorArr ^ arr[i];
 
        // If the XOR of the remaining elements
        // is equal to the current element
        if (arr[i] == x)
            return true;
    }
 
    // If no such element is found
    return false;
}
 
// Driver code
int main()
{
    int arr[] = { 8, 2, 4, 15, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    if (containsElement(arr, n))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}

Java




// Java implementation of the approach
class GFG
{
     
// Function that returns true if the array
// contains an element which is equal to
// the XOR of the remaining elements
static boolean containsElement(int [] arr, int n)
{
 
    // To store the XOR of all
    // the array elements
    int xorArr = 0;
    for (int i = 0; i < n; ++i)
        xorArr ^= arr[i];
 
    // For every element of the array
    for (int i = 0; i < n; ++i)
    {
 
        // Take the XOR after excluding
        // the current element
        int x = xorArr ^ arr[i];
 
        // If the XOR of the remaining elements
        // is equal to the current element
        if (arr[i] == x)
            return true;
    }
 
    // If no such element is found
    return false;
}
 
// Driver code
public static void main (String[] args)
{
    int [] arr = { 8, 2, 4, 15, 1 };
    int n = arr.length;
 
    if (containsElement(arr, n))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}
 
// This code is contributed by ihritik

Python3




# Python3 implementation of the approach
 
# Function that returns true if the array
# contains an element which is equal to
# the XOR of the remaining elements
def containsElement(arr, n):
 
    # To store the XOR of all
    # the array elements
    xorArr = 0
    for i in range(n):
        xorArr ^= arr[i]
 
    # For every element of the array
    for i in range(n):
 
        # Take the XOR after excluding
        # the current element
        x = xorArr ^ arr[i]
 
        # If the XOR of the remaining elements
        # is equal to the current element
        if (arr[i] == x):
            return True
 
    # If no such element is found
    return False
 
# Driver Code
arr = [8, 2, 4, 15, 1]
n = len(arr)
 
if (containsElement(arr, n)):
    print("Yes")
else:
    print("No")
 
# This code is contributed by Mohit Kumar

C#




// C# implementation of the approach
using System;
class GFG
{
     
// Function that returns true if the array
// contains an element which is equal to
// the XOR of the remaining elements
static bool containsElement(int [] arr, int n)
{
 
    // To store the XOR of all
    // the array elements
    int xorArr = 0;
    for (int i = 0; i < n; ++i)
        xorArr ^= arr[i];
 
    // For every element of the array
    for (int i = 0; i < n; ++i)
    {
 
        // Take the XOR after excluding
        // the current element
        int x = xorArr ^ arr[i];
 
        // If the XOR of the remaining elements
        // is equal to the current element
        if (arr[i] == x)
            return true;
    }
 
    // If no such element is found
    return false;
}
 
// Driver code
public static void Main ()
{
    int [] arr = { 8, 2, 4, 15, 1 };
    int n = arr.Length;
 
    if (containsElement(arr, n))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}
 
// This code is contributed by ihritik

Javascript




<script>
 
// JavaScript implementation of the approach
 
// Function that returns true if the array
// contains an element which is equal to
// the XOR of the remaining elements
function containsElement(arr, n)
{
 
    // To store the XOR of all
    // the array elements
    let xorArr = 0;
    for (let i = 0; i < n; ++i)
        xorArr ^= arr[i];
 
    // For every element of the array
    for (let i = 0; i < n; ++i) {
 
        // Take the XOR after excluding
        // the current element
        let x = xorArr ^ arr[i];
 
        // If the XOR of the remaining elements
        // is equal to the current element
        if (arr[i] == x)
            return true;
    }
 
    // If no such element is found
    return false;
}
 
// Driver code
    let arr = [ 8, 2, 4, 15, 1 ];
    let n = arr.length;
 
    if (containsElement(arr, n))
        document.write("Yes");
    else
        document.write("No");
 
// This code is contributed by Surbhi Tyagi.
 
</script>

Output: 

Yes

 

Time complexity: O(n) where n is size of input array

Auxiliary Space: O(1)


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