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

• Difficulty Level : Medium
• Last Updated : 06 Sep, 2022

Given an array of integers of size ‘n’ and an integer ‘k’,
We can perform the Bitwise AND 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 : k = 6 ; Array : 1, 2, 1, 2
Output : 0
Explanation : There are already two equal elements in the array so the answer is 0.
Input : k = 2 ; Array : 5, 6, 2, 4
Output : 1
Explanation : If we apply AND operation on element ‘6’, it will become 6&2 = 2
And the array will become 5 2 2 4,
Now, the array has two equal elements, so the answer is 1.
Input : k = 15 ; Array : 1, 2, 3
Output : -1
Explanation : No matter how many times we perform the above mentioned operation,
this array will never have equal element pair. So the answer is -1

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, no matter how many times you perform (y&k)
it’ll always give ‘y’ as the result.

• The answer will be ‘0’, if there are already equal elements in the array.
• 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’.
• 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’.
• If any of the above conditions are 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 count the// minimum operations required.int 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 AND 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 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 AND operations    return -1;} // Driver codeint main(){     int K = 3;    int a[] = { 1, 2, 3, 7 };    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.*; class geeks{     // Function to count the    // minimum operations required.    public 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            // try-catch is used so that            // nullpointer exception can be handled            try            {                if (map.get(a[i]))                    return 1;            }            catch (Exception e)            {                //TODO: handle exception            }                         try            {                map.put(a[i], true);            } catch (Exception e) {}        }         // create new array with AND 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 operation between            // 'k' and a[i] gives            // a number other than a[i]            if (a[i] != b[i])            {                try                {                    map.put(b[i], true);                }                catch (Exception e) {}            }        }         // 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            try            {                if (map.get(a[i]))                    return 1;            }            catch (Exception e)            {                //TODO: handle exception            }        }         // 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            try            {                if (map.get(b[i]))                    return 2;            }            catch (Exception e)            {                //TODO: handle exception            }             try            {                map.put(b[i], true);            }            catch (Exception e)            {                //TODO: handle exception            }        }         // otherwise it is impossible to        // create such an array with        // Bitwise AND operations        return -1;    }     // Driver Code    public static void main(String[] args)    {        int K = 3;        int[] a = { 1, 2, 3, 7 };        int n = a.length;         // Function call to compute the result        System.out.println(minOperations(a, n, K));    }} // This code is contributed by// sanjeev2552

## Python3

 # Python3 implementation of the approachfrom collections import defaultdict # Function to count the minimum# operations required.def minOperations(a, n, K):     Map = defaultdict(lambda:False)    for i in range(0, n):         # check if the initial array        # already contains an equal pair        if Map[a[i]] == True:            return 0        Map[a[i]] = True         # create new array with AND operations    b = []    for i in range(0, n):        b.append(a[i] & K)     # clear the map    Map.clear()     # Check if the solution    # is a single operation    for i in range(0, n):             # If Bitwise 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(0, n):         # Single operation        # will be enough        if Map[a[i]] == True:            return 1     # clear the map    Map.clear()     # Check if the solution    # is two operations    for i in range(0, n):             # Check if the array 'b'        # contains duplicates        if Map[b[i]] == True:            return 2         Map[b[i]] = True         # otherwise it is impossible to    # create such an array with    # Bitwise AND operations    return -1 # Driver codeif __name__ == "__main__":     K = 3    a = [1, 2, 3, 7]    n = len(a)     # Function call to compute the result    print(minOperations(a, n, K)) # This code is contributed by Rituraj Jain

## 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 AND 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, 2, 3, 7 };        int n = a.Length;        Console.WriteLine(minOperations(a, n, K));    }} // This code is contributed by Rajput-Ji

## Javascript



Output:

1

Time Complexity: O(n)

Auxiliary space: O(n) it is using extra space for unordere_map and array b

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