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

Last Updated : 05 Sep, 2022

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# 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)

Make Array elements equal by replacing adjacent elements with their XOR

NikhilRathor

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

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