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

• Difficulty Level : Medium
• Last Updated : 07 Sep, 2021

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<``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 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 code``int` `main()``{` `    ``int` `K = 3;``    ``int` `a[] = { 1, 2, 3, 7 };``    ``int` `n = ``sizeof``(a)/``sizeof``(a);` `    ``// Function call to compute the result``    ``cout << minOperations(a, n, K);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `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 approach``from` `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 code``if` `__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 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 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`

Complexity: O(n)

My Personal Notes arrow_drop_up