Find a number X such that XOR of given Array after adding X to each element is 0

Given an array arr[] of odd length N containing positive integers. The task is to find a positive integer X such that, adding X to all the elements of arr[] and then taking XOR of all the elements gives 0. Return -1 if no such X exists.

Examples: 

Input: arr[] = {2, 4, 5}
Output: 1
Explanation: Following are the operations performed in arr[] to get the desired result.
Adding 1 to each element in arr[] updates arr[] to arr[] = {3, 5, 6}
Now XOR of all the elements in arr[] is 3^5^6 = 0. 
Therefore, 1 is the required answer.  

Input: arr[] = {4, 5, 13}
Output: -1
Explanation: No such x exists for fulfilling the desired conditions. 

Approach: XOR of Odd number of 1's = 1, while even number of 1's = 0. This idea can be used to solve the given problem. Follow the steps mentioned below:

• Initialize variable X = 0.
• The binary representations of the array elements will be used for traversal of the elements and determining X.
• Start traversing from the 0th bit to 63rd bit.
• If at any bit position total number of set bits (1's) of the array elements are odd, add that power of 2 with X.
• If after completion of iteration there is odd number of 1 at any bit position then no such X exists. Otherwise, print X as answer.

See the illustrations below:

Illustration:

Case-1 (X possible): Take arr[] = { 2, 4, 5} 

	5th	4th	3rd	2nd	1st	0th
arr[0]	0	0	0	0	1	0
arr[1]	0	0	0	1	0	0
arr[2]	0	0	0	1	0	1
X	0	0	0	0	0	0

• Initially, X = 0 0 0 0 0 0, at 0th position set(1) bits are odd, so in order to make the set bits even,  flip the bits at 0th position. So, for flipping the bits just add (0 0 0 0 0 1) to all the elements of arr[] and in X.
• Now, the table will look like the following:

	5th	4th	3rd	2nd	1st	0th
arr[0]	0	0	0	0	1	1
arr[1]	0	0	0	1	0	1
arr[2]	0	0	0	1	1	0
XOR	0	0	0	0	0	0
X	0	0	0	0	0	1

• Now, the XOR of (arr[0]+x) ^  ( arr[1]+x) ^ arr[2]+x) = 0, result will be 0. So, print  res = X.

Take the following when no possible X

Case-2: Take example: arr[] = { 4, 5, 13 } 

	5th	4th	3rd	2nd	1st	0th
arr[0]	0	0	0	1	0	0
arr[1]	0	0	0	1	0	1
arr[2]	0	0	1	1	0	1
XOR	0	0	1	1	0	0
X	0	0	0	0	0	0

          XOR = Arr[0] ^ Arr[1] ^ Arr[2] = 1 1 0 0,   Here there are odd number of 1's at 2nd and 3rd bits.

• So, add 2pow(2nd) to all elements of arr and in X, then again will take the XOR, after this the elements become:

	5th	4th	3rd	2nd	1st	0th
arr[0]	0	0	1	0	0	0
arr[1]	0	0	1	0	0	1
arr[2]	0	1	0	0	0	1
XOR	0	1	0	0	0	0
X	0	0	0	1	0	0

If this keeps on going the left most 1 in XOR keeps on moving left.

Below is the implementation of the above approach:

// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find required result
long long solve(vector<long long>& a,
                int n)
{
    long long res = 0, j = 0, one = 1;

    // For 64 Bit
    while (j < 64) {
        // j is traversing each bit
        long long Xor = 0;
        long long powerOf2 = one << j;

        for (auto x : a)
            Xor ^= x;

        if (j == 63 && (Xor & powerOf2))
            return -1;

        if (Xor & powerOf2) {
            res += powerOf2;
            for (int i = 0; i < n; i++)
                a[i] += powerOf2;
        }
        j++;
    }
    return res;
}

// Driver Code
int main()
{

    // Size of arr[]
    int N = 3;
    vector<long long> arr = { 2, 4, 5 };

    cout << solve(arr, N) << '\n';

    return 0;
}

// Java program for above approach
class GFG{

// Function to find required result
static long solve(int[] a,
                int n)
{
    long res = 0, j = 0, one = 1;

    // For 64 Bit
    while (j < 64) 
    {
      
        // j is traversing each bit
        long Xor = 0;
        long powerOf2 = one << j;

        for (int x : a)
            Xor ^= x;

        if (j == 63 && (Xor & powerOf2)!=0)
            return -1;

        if ((Xor & powerOf2)!=0) {
            res += powerOf2;
            for (int i = 0; i < n; i++)
                a[i] += powerOf2;
        }
        j++;
    }
    return res;
}

// Driver Code
public static void main(String[] args)
{

    // Size of arr[]
    int N = 3;
    int[] arr = { 2, 4, 5 };

    System.out.print(solve(arr, N));
}
}

// This code is contributed by shikhasingrajput 

# python program for above approach

# Function to find required result
def solve(a, n):

    res = 0
    j = 0
    one = 1

    # For 64 Bit
    while (j < 64):
                # j is traversing each bit
        Xor = 0
        powerOf2 = one << j

        for x in a:
            Xor ^= x

        if (j == 63 and (Xor & powerOf2)):
            return -1

        if (Xor & powerOf2):
            res += powerOf2
            for i in range(0, n):
                a[i] += powerOf2

        j += 1

    return res

# Driver Code
if __name__ == "__main__":

        # Size of arr[]
    N = 3
    arr = [2, 4, 5]

    print(solve(arr, N))

    # This code is contributed by rakeshsahni

// C# program for above approach
using System;
class GFG
{

    // Function to find required result
    static long solve(int[] a, int n)
    {
        int res = 0, j = 0, one = 1;

        // For 64 Bit
        while (j < 64)
        {

            // j is traversing each bit
            long Xor = 0;
            long powerOf2 = one << j;

            foreach (int x in a)
                Xor ^= x;

            if (j == 63 && (Xor & powerOf2) != 0)
                return -1;

            if ((Xor & powerOf2) != 0)
            {
                res += (int)powerOf2;
                for (int i = 0; i < n; i++)
                    a[i] += (int)powerOf2;
            }
            j++;
        }
        return res;
    }

    // Driver Code
    public static void Main()
    {

        // Size of arr[]
        int N = 3;
        int[] arr = { 2, 4, 5 };

        Console.Write(solve(arr, N));
    }
}

// This code is contributed by gfgking 

<script>

// JavaScript program for above approach 

// Function to find required result
function solve(a, n)
{
    let res = 0, j = 0, one = 1;
    
    // For 64 Bit
    while (j < 64)
    {
        
        // j is traversing each bit
        let Xor = 0;
        let powerOf2 = one << j;
    
        for(let x of a)
            Xor ^= x;
    
        if (j == 63 && (Xor & powerOf2))
            return -1;
    
        if (Xor & powerOf2) 
        {
            res += powerOf2;
            for(let i = 0; i < n; i++)
                a[i] += powerOf2;
        }
        j++;
    }
    return res;
}

// Driver Code

// Size of arr[]
let N = 3;
let arr = [ 2, 4, 5 ];

document.write(solve(arr, N) + '<br>');

// This code is contributed by Potta Lokesh

</script>

1

Time Complexity: O(N*logN)
Auxiliary Space: O(1)

madhav_mohan

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Article Tags :

• Bit Magic
• Mathematical
• Competitive Programming
• DSA
• Arrays
• Bitwise-XOR
• Bit Algorithms

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Practice Tags :

• Arrays
• Bit Magic
• Mathematical