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Minimum Bitwise OR operations to make any two array elements equal

  • Last Updated : 08 Jun, 2021

Given an array arr[] of integers and an integer K, we can perform the Bitwise OR 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:
We can OR a[0] with x, which makes it 3 which is equal to a[3]
Input : arr[] = {13, 26, 21, 15}, K = 13 
Output : -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 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 OR 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 OR 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 OR 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
    public 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]))
                return 0;
 
            map.put(a[i], true);
        }
 
        // Create new array with OR 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.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]))
                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.put(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;
        System.out.println(minOperations(a, n, K));
    }
}
 
// This code is contributed by
// sanjeev2552

Python3




# Python3 implementation of the approach
 
# Function to return the count of
# minimum operations required
def minOperations(a, n, K) :
     
    map = dict.fromkeys(a,0) ;
     
    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 OR 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 OR 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] not in map :
            pass
         
        elif (map[a[i]]) :
            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 (map[b[i]]) :
            return 2;
 
        map[b[i]] = true;
 
    # Otherwise it is impossible to
    # create such an array with
    # Bitwise OR 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 OR 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 29AjayKumar

Javascript




<script>
 
// JavaScript implementation of the approach
 
// Function to return the count of
// minimum operations required
function minOperations(a, n, K) {
    let map = new Map();
    for (let i = 0; i < n; i++) {
 
        // Check if the initial array
        // already contains an equal pair
        if (map.has(a[i]))
            return 0;
        map.set(a[i], true);
    }
 
    // Create new array with OR operations
    let b = new Array(n);
    for (let i = 0; i < n; i++)
        b[i] = a[i] | K;
 
    // Clear the map
    map.clear();
 
    // Check if the solution
    // is a single operation
    for (let 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.set(b[i], true);
    }
 
    // Check if any of the a[i]
    // gets equal to any other element
    // of the array after the operation
    for (let i = 0; i < n; i++)
 
        // Single operation
        // will be enough
        if (map.has(a[i]))
            return 1;
 
    // Clear the map
    map.clear();
 
    // Check if the solution
    // is two operations
    for (let i = 0; i < n; i++) {
 
        // Check if the array 'b'
        // contains duplicates
        if (map.has(b[i]))
            return 2;
 
        map.set(b[i], true);
    }
 
    // Otherwise it is impossible to
    // create such an array with
    // Bitwise OR operations
    return -1;
}
 
// Driver code
 
let K = 3;
let a = [1, 9, 4, 3];
let n = a.length;
 
// Function call to compute the result
document.write(minOperations(a, n, K));
 
// This code is contributed by gfgking
 
</script>
Output: 
1

 

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