# Minimum Bitwise XOR operations to make any two array elements equal

• Last Updated : 05 Sep, 2022

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 using namespace std; // Function to return the count of// minimum operations requiredint minOperations(int a[], int n, int K){    unordered_map 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 codeint 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 approachimport java.util.HashMap; class GFG{    // Function to return the count of    // minimum operations required    static int minOperations(int[] a, int n, int k)    {        HashMap 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 requireddef 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 codeif __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 approachusing 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 map =                new Dictionary();         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



Output:

-1

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

Auxiliary space: O(n)

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