Skip to content
Related Articles

Related Articles

Improve Article

Minimum Bitwise XOR operations to make any two array elements equal

  • Last Updated : 10 May, 2021
Geek Week

Given an array arr[] of integers of size N and an integer K. One can perform the Bitwise XOR operation between any array element and K any number of times. The task is to print the minimum number of such operations required to make any two elements of the array equal. If it is not possible to make any two elements of the array equal after performing the above-mentioned operation then print -1.
Examples: 
 

Input : arr[] = {1, 9, 4, 3}, K = 3 
Output :-1 
Explanation : No possible to make any two elements equal
Input : arr[] = {13, 13, 21, 15}, K = 13 
Output :
Explanation : Already exists two same elements

 

Approach: The key observation is that if it is possible to make the desired array then the answer will be either 0, 1 or 2. It will never exceed 2. 
 

Because, if (x ^ k) = y 
then, performing (y ^ k) will give x again 
 



 

  1. The answer will be 0, if there are already equal elements in the array.
  2. For the answer to be 1, we will create a new array b[] which holds b[i] = (a[i] ^ K)
    Now, for each a[i] we will check if there is any index j such that i != j and a[i] = b[j].
    If yes, then the answer will be 1.
  3. For the answer to be 2, we will check for an index i in the new array b[], If there is any index j such that i != j and b[i] = b[j]
    If yes, then the answer will be 2.
  4. If any of the above conditions is not satisfied then the answer will be -1.

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 count of
// minimum operations required
int minOperations(int a[], int n, int K)
{
    unordered_map<int, bool> map;
    for (int i = 0; i < n; i++) {
 
        // Check if the initial array
        // already contains an equal pair
        if (map[a[i]])
            return 0;
        map[a[i]] = true;
    }
 
    // Create new array with XOR operations
    int b[n];
    for (int i = 0; i < n; i++)
        b[i] = a[i] ^ K;
 
    // Clear the map
    map.clear();
 
    // Check if the solution
    // is a single operation
    for (int i = 0; i < n; i++) {
 
        // If Bitwise XOR operation between
        // 'k' and a[i] gives
        // a number other than a[i]
        if (a[i] != b[i])
            map[b[i]] = true;
    }
 
    // Check if any of the a[i]
    // gets equal to any other element
    // of the array after the operation
    for (int i = 0; i < n; i++)
 
        // Single operation
        // will be enough
        if (map[a[i]])
            return 1;
 
    // Clear the map
    map.clear();
 
    // Check if the solution
    // is two operations
    for (int i = 0; i < n; i++) {
 
        // Check if the array 'b'
        // contains duplicates
        if (map[b[i]])
            return 2;
 
        map[b[i]] = true;
    }
 
    // Otherwise it is impossible to
    // create such an array with
    // Bitwise XOR operations
    return -1;
}
 
// Driver code
int main()
{
 
    int K = 3;
    int a[] = { 1, 9, 4, 3 };
    int n = sizeof(a) / sizeof(a[0]);
 
    // Function call to compute the result
    cout << minOperations(a, n, K);
 
    return 0;
}

Java




// Java implementation of the approach
import java.util.HashMap;
 
class GFG
{
    // Function to return the count of
    // minimum operations required
    static int minOperations(int[] a, int n, int k)
    {
        HashMap<Integer,
                Boolean> map = new HashMap<>();
        for (int i = 0; i < n; i++)
        {
 
            // Check if the initial array
            // already contains an equal pair
            if (map.containsKey(a[i]) &&
                map.get(a[i]))
                return 0;
            map.put(a[i], true);
        }
 
        // Create new array with XOR operations
        int[] b = new int[n];
        for (int i = 0; i < n; i++)
            b[i] = a[i] ^ k;
 
        // Clear the map
        map.clear();
 
        // Check if the solution
        // is a single operation
        for (int i = 0; i < n; i++)
        {
 
            // If Bitwise XOR operation between
            // 'k' and a[i] gives
            // a number other than a[i]
            if (a[i] != b[i])
                map.put(b[i], true);
        }
 
        // Check if any of the a[i]
        // gets equal to any other element
        // of the array after the operation
        for (int i = 0; i < n; i++)
         
            // Single operation
            // will be enough
            if (map.containsKey(a[i]) &&
                map.get(a[i]))
                return 1;
 
        // Clear the map
        map.clear();
 
        // Check if the solution
        // is two operations
        for (int i = 0; i < n; i++)
        {
 
            // Check if the array 'b'
            // contains duplicates
            if (map.containsKey(b[i]) &&
                map.get(b[i]))
                return 2;
 
            map.put(b[i], true);
        }
 
        // Otherwise it is impossible to
        // create such an array with
        // Bitwise XOR operations
        return -1;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int K = 3;
        int[] a = { 1, 9, 4, 3 };
        int n = a.length;
        System.out.println(minOperations(a, n, K));
    }
}
 
// This code is contributed by
// Vivek Kumar Singh

Python3




# Python3 implementation of the approach
 
# Function to return the count of
# minimum operations required
def minOperations(a, n, K) :
 
    map = dict.fromkeys(a, False);
    for i in range(n) :
 
        # Check if the initial array
        # already contains an equal pair
        if (map[a[i]]) :
            return 0;
        map[a[i]] = True;
 
    # Create new array with XOR operations
    b = [0] * n;
    for i in range(n) :
        b[i] = a[i] ^ K;
 
    # Clear the map
    map.clear();
 
    # Check if the solution
    # is a single operation
    for i in range(n) :
 
        # If Bitwise XOR operation between
        # 'k' and a[i] gives
        # a number other than a[i]
        if (a[i] != b[i]) :
            map[b[i]] = True;
     
    # Check if any of the a[i]
    # gets equal to any other element
    # of the array after the operation
    for i in range(n) :
 
        # Single operation
        # will be enough
        if a[i] in map :
            return 1;
 
    # Clear the map
    map.clear();
 
    # Check if the solution
    # is two operations
    for i in range(n) :
         
        # Check if the array 'b'
        # contains duplicates
        if b[i] in map :
            return 2;
 
        map[b[i]] = True;
 
    # Otherwise it is impossible to
    # create such an array with
    # Bitwise XOR operations
    return -1;
 
# Driver code
if __name__ == "__main__" :
 
    K = 3;
    a = [ 1, 9, 4, 3 ];
    n = len(a);
 
    # Function call to compute the result
    print(minOperations(a, n, K));
 
# This code is contributed by AnkitRai01

C#




// C# implementation of the approach
using System;
using System.Collections.Generic;
 
class GFG
{
     
    // Function to return the count of
    // minimum operations required
    public static int minOperations(int[] a,
                                    int n, int K)
    {
 
        Dictionary<int, Boolean> map =
                new Dictionary<int, Boolean>();
 
        for (int i = 0; i < n; i++)
        {
 
            // Check if the initial array
            // already contains an equal pair
            if (map.ContainsKey(a[i]))
                return 0;
 
            map.Add(a[i], true);
        }
 
        // Create new array with XOR operations
        int[] b = new int[n];
        for (int i = 0; i < n; i++)
            b[i] = a[i] ^ K;
 
        // Clear the map
        map.Clear();
 
        // Check if the solution
        // is a single operation
        for (int i = 0; i < n; i++)
        {
 
            // If Bitwise OR operation between
            // 'k' and a[i] gives
            // a number other than a[i]
            if (a[i] != b[i])
                map.Add(b[i], true);
        }
 
        // Check if any of the a[i]
        // gets equal to any other element
        // of the array after the operation
        for (int i = 0; i < n; i++)
        {
 
            // Single operation
            // will be enough
            if (map.ContainsKey(a[i]))
                return 1;
        }
 
        // Clear the map
        map.Clear();
 
        // Check if the solution
        // is two operations
        for (int i = 0; i < n; i++)
        {
 
            // Check if the array 'b'
            // contains duplicates
            if (map.ContainsKey(b[i]))
                return 2;
            map.Add(b[i], true);
        }
 
        // Otherwise it is impossible to
        // create such an array with
        // Bitwise OR operations
        return -1;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int K = 3;
        int[] a = { 1, 9, 4, 3 };
        int n = a.Length;
        Console.WriteLine(minOperations(a, n, K));
    }
}
 
// This code is contributed by Rajput-Ji

Javascript




<script>
 
 
// Javascript implementation of the approach
 
// Function to return the count of
// minimum operations required
function minOperations(a, n, K)
{
    var map = new Map();
    for (var i = 0; i < n; i++) {
 
        // Check if the initial array
        // already contains an equal pair
        if (map[a[i]])
            return 0;
        map[a[i]] = true;
    }
 
    // Create new array with XOR operations
    var b = Array(n);
    for (var i = 0; i < n; i++)
        b[i] = a[i] ^ K;
 
    // Clear the map
    map = new Map();
 
    // Check if the solution
    // is a single operation
    for (var i = 0; i < n; i++) {
 
        // If Bitwise XOR operation between
        // 'k' and a[i] gives
        // a number other than a[i]
        if (a[i] != b[i])
            map[b[i]] = true;
    }
 
    // Check if any of the a[i]
    // gets equal to any other element
    // of the array after the operation
    for (var i = 0; i < n; i++)
 
        // Single operation
        // will be enough
        if (map[a[i]])
            return 1;
 
    // Clear the map
    map = new Map();
 
    // Check if the solution
    // is two operations
    for (var i = 0; i < n; i++) {
 
        // Check if the array 'b'
        // contains duplicates
        if (map[b[i]])
            return 2;
 
        map[b[i]] = true;
    }
 
    // Otherwise it is impossible to
    // create such an array with
    // Bitwise XOR operations
    return -1;
}
 
// Driver code
var K = 3;
var a = [ 1, 9, 4, 3 ];
var n = a.length;
// Function call to compute the result
document.write( minOperations(a, n, K));
 
</script>   
Output: 
-1

 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :