Maximize Bitwise AND of first element with complement of remaining elements for any permutation of given Array
Given an array arr[] consisting of N integers, the task is to find the maximum value of Bitwise AND of the first element with the complement of remaining elements for any permutation of this array, i.e.
A1 &(~A2) & (~A3) & ……& (~An)
Examples:
Input: arr[] = {1, 2, 4, 8, 16}
Output: 16
Explanation:
For the permutation {16, 1, 2, 4, 8}, the maximum value of the expression can be obtained.
Input: arr[] = {0, 2, 3, 4, 9, 8}
Output: 4
Explanation:
For the permutation {4, 8, 9, 3, 2, 0}, the maximum value of the expression can be obtained
Naive Approach: The simplest approach to solve the problem is to generate all possible permutations of the given array and find the required value for each permutation and print the maximum among them.
Time Complexity: O(N * N!)
Auxiliary Space: O(N)
Efficient Approach: The above approach can be optimized by the following observation:
- The expression A1 &(~A2) & (~A3) & …… & (~An) is solely dependent on the value of A1.
- Therefore, to get the maximize the value from the expression, choose A1 such that it has the set bit of as highest significance as possible, which is unset in all other array elements.
- Since, the order of remaining array elements doesn’t matter, print any permutation having the obtained A1 as the first element.
Illustration:
For arr[] = {1, 2, 4, 8, 16}
Binary representation of the array elements:
(16)10 = (10000)2
(8)10 = (01000)2
(4)10 = (00100)2
(2)10 = (00010)2
(1)10 = (00001)2
As it can be seen that 16 has the highest significant set bit which is unset in all other array elements. Therefore, the required permutation of given permutation will contain 16 as the first element.
Hence, the required Bitwise AND is maximum for permutation having 16 as the first element, which is equal to 16 in this case.
Below is the implementation of the above approach:
C++
// C++ Program to implement // the above approach #include <bits/stdc++.h> using namespace std; #define size_int 32 // Function to maximize the value for // the given function and the array elements int functionMax( int arr[], int n) { // Vector array to maintain which bit is set // for which integer in the given array by // saving index of that integer vector< int > setBit[32]; for ( int i = 0; i < n; i++) { for ( int j = 0; j < size_int; j++) { // Check if j-th bit is set for // i-th integer if (arr[i] & (1 << j)) // Push the index of that // integer in setBit[j] setBit[j].push_back(i); } } // Find the element having // highest significant set bit // unset in other elements for ( int i = size_int; i >= 0; i--) { if (setBit[i].size() == 1) { // Place that integer at 0-th index swap(arr[0], arr[setBit[i][0]]); break ; } } // Store the maximum AND value int maxAnd = arr[0]; for ( int i = 1; i < n; i++) { maxAnd = maxAnd & (~arr[i]); } // Return the answer return maxAnd; } // Driver Code int main() { int arr[] = { 1, 2, 4, 8, 16 }; int n = sizeof arr / sizeof arr[0]; // Function call cout << functionMax(arr, n); return 0; } |
Java
// Java Program to implement // the above approach import java.util.*; class GFG{ static final int size_int = 32 ; // Function to maximize the value for // the given function and the array elements static int functionMax( int arr[], int n) { // Vector array to maintain which bit is set // for which integer in the given array by // saving index of that integer Vector<Integer> []setBit = new Vector[ 32 + 1 ]; for ( int i = 0 ; i < setBit.length; i++) setBit[i] = new Vector<Integer>(); for ( int i = 0 ; i < n; i++) { for ( int j = 0 ; j < size_int; j++) { // Check if j-th bit is set for // i-th integer if ((arr[i] & ( 1 << j)) > 0 ) // Push the index of that // integer in setBit[j] setBit[j].add(i); } } // Find the element having // highest significant set bit // unset in other elements for ( int i = size_int; i >= 0 ; i--) { if (setBit[i].size() == 1 ) { // Place that integer at 0-th index swap(arr, 0 , setBit[i].get( 0 )); break ; } } // Store the maximum AND value int maxAnd = arr[ 0 ]; for ( int i = 1 ; i < n; i++) { maxAnd = maxAnd & (~arr[i]); } // Return the answer return maxAnd; } static int [] swap( int []arr, int i, int j) { int temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; return arr; } // Driver Code public static void main(String[] args) { int arr[] = { 1 , 2 , 4 , 8 , 16 }; int n = arr.length; // Function call System.out.print(functionMax(arr, n)); } } // This code is contributed by PrinciRaj1992 |
Python3
# Python 3 Program to # implement the above approach # Function to maximize the # value for the given function # and the array elements def functionMax(arr, n): # Vector array to maintain # which bit is set for which # integer in the given array by # saving index of that integer setBit = [[] for i in range ( 32 )] for i in range (n): for j in range ( 32 ): # Check if j-th bit is # set for i-th integer if (arr[i] & ( 1 << j)): # Push the index of that # integer in setBit[j] setBit[j].append(i) # Find the element having # highest significant set bit # unset in other elements i = 31 while (i > = 0 ): if ( len (setBit[i]) = = 1 ): # Place that integer # at 0-th index temp = arr[ 0 ] arr[ 0 ] = arr[setBit[i][ 0 ]] arr[setBit[i][ 0 ]] = temp break i - = 1 # Store the maximum # AND value maxAnd = arr[ 0 ] for i in range ( 1 , n, 1 ): maxAnd = (maxAnd & (~arr[i])) # Return the answer return maxAnd # Driver Code if __name__ = = '__main__' : arr = [ 1 , 2 , 4 , 8 , 16 ] n = len (arr) # Function call print (functionMax(arr, n)) # This code is contributed by bgangwar59 |
C#
// C# Program to implement // the above approach using System; using System.Collections.Generic; class GFG{ static readonly int size_int = 32; // Function to maximize the value for // the given function and the array elements static int functionMax( int []arr, int n) { // List array to maintain which bit is set // for which integer in the given array by // saving index of that integer List< int > []setBit = new List< int >[32 + 1]; for ( int i = 0; i < setBit.Length; i++) setBit[i] = new List< int >(); for ( int i = 0; i < n; i++) { for ( int j = 0; j < size_int; j++) { // Check if j-th bit is set for // i-th integer if ((arr[i] & (1 << j)) > 0) // Push the index of that // integer in setBit[j] setBit[j].Add(i); } } // Find the element having // highest significant set bit // unset in other elements for ( int i = size_int; i >= 0; i--) { if (setBit[i].Count == 1) { // Place that integer at 0-th index swap(arr, 0, setBit[i][0]); break ; } } // Store the maximum AND value int maxAnd = arr[0]; for ( int i = 1; i < n; i++) { maxAnd = maxAnd & (~arr[i]); } // Return the answer return maxAnd; } static int [] swap( int []arr, int i, int j) { int temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; return arr; } // Driver Code public static void Main(String[] args) { int []arr = { 1, 2, 4, 8, 16 }; int n = arr.Length; // Function call Console.Write(functionMax(arr, n)); } } // This code is contributed by Rajput-Ji |
Javascript
<script> // Javascript Program to implement // the above approach var size_int = 32; // Function to maximize the value for // the given function and the array elements function functionMax( arr, n) { // Vector array to maintain which bit is set // for which integer in the given array by // saving index of that integer var setBit = Array.from(Array(32), ()=> new Array()); for ( var i = 0; i < n; i++) { for ( var j = 0; j < size_int; j++) { // Check if j-th bit is set for // i-th integer if (arr[i] & (1 << j)) // Push the index of that // integer in setBit[j] setBit[j].push(i); } } // Find the element having // highest significant set bit // unset in other elements for ( var i = size_int-1; i >= 0; i--) { if (setBit[i].length == 1) { // Place that integer at 0-th index [arr[0], arr[setBit[i][0]]] = [arr[setBit[i][0]], arr[0]]; break ; } } // Store the maximum AND value var maxAnd = arr[0]; for ( var i = 1; i < n; i++) { maxAnd = maxAnd & (~arr[i]); } // Return the answer return maxAnd; } // Driver Code var arr = [1, 2, 4, 8, 16]; var n = arr.length; // Function call document.write( functionMax(arr, n)); // This code is contributed by rrrtnx. </script> |
16
Time Complexity: O(N * sizeof(int)), where sizeof(int) is 32
Auxiliary Space: O(N)