Minimum Bitwise AND operations to make any two array elements equal
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 <bits/stdc++.h> 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[0]); // Function call to compute the result cout << minOperations(a, n, K); return 0; } // 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<Integer, Boolean> 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 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# 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 <script> // JavaScript implementation of the approach // Function to count the // 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.set(a[i], true); } // create new array with AND 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 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 (var i = 0; i < n; i++) // Single operation // will be enough if (map.get(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.get(b[i])) return 2; map.set(b[i], true); } // otherwise it is impossible to // create such an array with // Bitwise AND operations return -1; } // Driver code var K = 3; var a = [1, 2, 3, 7]; var n = a.length; // Function call to compute the result document.write( minOperations(a, n, K)); </script> Time Complexity: O(n)
Auxiliary space: O(n) it is using extra space for unordere_map and array b
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