Rearrange an array to maximize sum of Bitwise AND of same-indexed elements with another array

Last Updated : 19 Oct, 2023

Given two arrays A[] and B[] of sizes N, the task is to find the maximum sum of Bitwise AND of same-indexed elements in the arrays A[] and B[] that can be obtained by rearranging the array B[] in any order.

Examples:

Input: A[] = {1, 2, 3, 4}, B[] = {3, 4, 1, 2}
Output: 10
Explanation: One possible way is to obtain the maximum value is to rearrange the array B[] to {1, 2, 3, 4}.
Therefore, sum of Bitwise AND of same-indexed elements of the arrays A[] and B[] = { 1&1 + 2&2 + 3&3 + 4&4 = 10), which is the maximum possible.

Input: A[] = {3, 5, 7, 11}, B[] = {2, 6, 10, 12}
Output: 22

Naive Approach: The simplest approach to solve the problem is to generate all possible permutations of array B[] and for each permutation, calculate the sum of Bitwise AND of same-indexed elements in arrays A[] and B[] and update the maximum possible sum accordingly. Finally, print the maximum sum possible.

Time Complexity: O(N! * N)
Auxiliary Space: O(1)

Efficient Approach: The above approach can also be optimized based on the following observations:

For each array element of A[] the idea is to chose a not selected array element of B[] using bitmasking which will give maximum bitwise AND sum upto the current index.
The idea is to use Dynamic Programming with bitmasking as it has overlapping subproblems and optimal substructure.
Suppose, dp(i, mask) represents the maximum bitwise AND sum of array A[] and i, with the selected elements of array B[] represented by bits-position of mask.
Then the transition from one state to another state can be defined as:
For all j in the range [0, N]:
If the j^th bit of mask is not set then, dp(i, mask) = max(dp(i, mask|(1<<j))).

Follow the steps below to solve the problem:

Define a vector of vectors, says dp of dimension N*2^N with value -1 to store all dp-states.
Define a recursive Dp function say maximizeAndUtil(i, mask) to find the maximum sum of the bitwise AND of the elements at the same respective positions in both arrays A[] and B[]:
- In the base case, if i is equal to N then return 0.
- If dp[i][mask] is not equal to -1 i.e already visited then return dp[i][mask].
- Iterate over the range [0, N-1] using variable j and in each iteration, If j^th bit in the mask is not set then update dp[i][mask] as dp[i][mask] = max(dp[i][mask], maximizeUtil(i+1, mask| 2^j).
- Finally, return dp[i][mask].
Call the recursive function maximizeAnd(0, 0) and print the value returned by it as the answer.

Below is the implementation of the above approach:

C++

// C++ program for the above approach #include <bits/stdc++.h> using namespace std;  // Function to implement recursive DP int maximizeAnd(int i, int mask,                 int* A, int* B, int N,                 vector<vector<int> >& dp) {     // If i is equal to N     if (i == N)         return 0;      // If dp[i][mask] is not     // equal to -1     if (dp[i][mask] != -1)         return dp[i][mask];      // Iterate over the array B[]     for (int j = 0; j < N; ++j) {          // If current element         // is not yet selected         if (!(mask & (1 << j))) {              // Update dp[i][mask]             dp[i][mask] = max(                 dp[i][mask],                 (A[i] & B[j])                     + maximizeAnd(i + 1, mask | (1 << j), A,                                   B, N, dp));         }     }     // Return dp[i][mask]     return dp[i][mask]; }  // Function to obtain maximum sum // of Bitwise AND of same-indexed // elements from the arrays A[] and B[] int maximizeAndUtil(int* A, int* B, int N) {     // Stores all dp-states     vector<vector<int> > dp(         N, vector<int>(1 << N + 1, -1));      // Returns the maximum value     // returned by the function maximizeAnd()     return maximizeAnd(0, 0, A, B, N, dp); }  // Driver Code int main() {     int A[] = { 3, 5, 7, 11 };     int B[] = { 2, 6, 10, 12 };     int N = sizeof A / sizeof A[0];      cout << maximizeAndUtil(A, B, N); }

Java

// Java program for the above approach import java.io.*; import java.lang.*; import java.util.*;  public class GFG {      // Function to implement recursive DP     static int maximizeAnd(int i, int mask, int A[],                            int B[], int N, int[][] dp)     {         // If i is equal to N         if (i == N)             return 0;          // If dp[i][mask] is not         // equal to -1         if (dp[i][mask] != -1)             return dp[i][mask];          // Iterate over the array B[]         for (int j = 0; j < N; ++j) {              // If current element             // is not yet selected             if ((mask & (1 << j)) == 0) {                  // Update dp[i][mask]                 dp[i][mask] = Math.max(                     dp[i][mask],                     (A[i] & B[j])                         + maximizeAnd(i + 1,                                       mask | (1 << j), A, B,                                       N, dp));             }         }         // Return dp[i][mask]         return dp[i][mask];     }      // Function to obtain maximum sum     // of Bitwise AND of same-indexed     // elements from the arrays A[] and B[]     static int maximizeAndUtil(int A[], int B[], int N)     {                // Stores all dp-states         int dp[][] = new int[N][(1 << N) + 1];         for (int dd[] : dp)             Arrays.fill(dd, -1);          // Returns the maximum value         // returned by the function maximizeAnd()         return maximizeAnd(0, 0, A, B, N, dp);     }      // Driver Code     public static void main(String[] args)     {         int A[] = { 3, 5, 7, 11 };         int B[] = { 2, 6, 10, 12 };         int N = A.length;          System.out.print(maximizeAndUtil(A, B, N));     } }  // This code is contributed by Kingash.

Python3

# Python3 program for the above approach  # Function to implement recursive DP def maximizeAnd(i, mask, A, B, N, dp):          # If i is equal to N     if (i == N):         return 0      # If dp[i][mask] is not     # equal to -1     if (dp[i][mask] != -1):         return dp[i][mask]      # Iterate over the array B[]     for j in range(N):                  # If current element         # is not yet selected         if ((mask & (1 << j)) == 0):                          # Update dp[i][mask]             dp[i][mask] = max(                 dp[i][mask],(A[i] & B[j]) +                  maximizeAnd(i + 1, mask | (1 << j),                             A, B, N, dp))                      # Return dp[i][mask]     return dp[i][mask]  # Function to obtain maximum sum # of Bitwise AND of same-indexed # elements from the arrays A[] and B[] def maximizeAndUtil(A, B, N):          # Stores all dp-states     temp = [-1 for i in range(1 << N + 1)]     dp = [temp for i in range(N)]      # Returns the maximum value     # returned by the function maximizeAnd()     return maximizeAnd(0, 0, A, B, N, dp)  # Driver Code if __name__ == '__main__':          A = [ 3, 5, 7, 11 ]     B = [ 2, 6, 10, 12 ]     N = len(A)      print(maximizeAndUtil(A, B, N))      # This code is contributed by ipg2016107

// C# program for the above approach using System;  class GFG {      // Function to implement recursive DP     static int maximizeAnd(int i, int mask, int[] A,                            int[] B, int N, int[,] dp)     {         // If i is equal to N         if (i == N)             return 0;          // If dp[i][mask] is not         // equal to -1         if (dp[i, mask] != -1)             return dp[i, mask];          // Iterate over the array B[]         for (int j = 0; j < N; ++j) {              // If current element             // is not yet selected             if ((mask & (1 << j)) == 0) {                  // Update dp[i][mask]                 dp[i, mask] = Math.Max(                     dp[i, mask],                     (A[i] & B[j])                         + maximizeAnd(i + 1,                                       mask | (1 << j), A, B,                                       N, dp));             }         }         // Return dp[i][mask]         return dp[i, mask];     }      // Function to obtain maximum sum     // of Bitwise AND of same-indexed     // elements from the arrays A[] and B[]     static int maximizeAndUtil(int[] A, int[] B, int N)     {                // Stores all dp-states         int[,] dp = new int[N, (1 << N) + 1];         for(int i = 0; i<N; i++)         {             for(int j =0 ; j<(1 << N) + 1; j++)             {                 dp[i, j] = -1;             }         }          // Returns the maximum value         // returned by the function maximizeAnd()         return maximizeAnd(0, 0, A, B, N, dp);     }      // Driver Code     static void Main()     {         int[] A = { 3, 5, 7, 11 };         int[] B = { 2, 6, 10, 12 };         int N = A.Length;          Console.Write(maximizeAndUtil(A, B, N));     } }  // This code is contributed by sanjoy_62.

JavaScript

<script>  // Javascript program for the above approach  // Function to implement recursive DP function maximizeAnd(i, mask, A, B, N, dp) {          // If i is equal to N     if (i == N)         return 0;      // If dp[i][mask] is not     // equal to -1     if (dp[i][mask] != -1)         return dp[i][mask];      // Iterate over the array B[]     for(var j = 0; j < N; ++j)     {                  // If current element         // is not yet selected         if (!(mask & (1 << j)))         {                          // Update dp[i][mask]             dp[i][mask] = Math.max(                 dp[i][mask], (A[i] & B[j]) +                  maximizeAnd(i + 1, mask | (1 << j), A,                             B, N, dp));         }     }          // Return dp[i][mask]     return dp[i][mask]; }  // Function to obtain maximum sum // of Bitwise AND of same-indexed // elements from the arrays A[] and B[] function maximizeAndUtil(A, B, N) {          // Stores all dp-states     var dp = Array.from(         Array(N), () => Array(1 << N + 1).fill(-1));      // Returns the maximum value     // returned by the function maximizeAnd()     return maximizeAnd(0, 0, A, B, N, dp); }  // Driver Code var A = [ 3, 5, 7, 11 ]; var B = [ 2, 6, 10, 12 ]; var N = A.length  document.write(maximizeAndUtil(A, B, N));  // This code is contributed by rrrtnx  </script>

Output

Time Complexity: O (N² * 2^N)
Auxiliary Space: O(N * 2^N)

DP Tabulation Approach(Iterative approach): The approach to solving this problem is the same but the DP tabulation(bottom-up) method is better than the Dp + memoization(top-down) because the memoization method needs extra stack space of recursion calls. Below are the steps:

Create a 2D matrix, say DP[][] to store the solution of the subproblems and initialize it with 0.
Initialize the DP with base cases.
Now Iterate over subproblems to get the value of the current problem from the previous computation of subproblems stored in the DP[][] as the transition from one state to another state can be defined as:
- For all j in the range [0, N], If the j^th bit of mask is not set then, dp(i, mask) = max(dp(i, mask|(1<<j))).
Return the final solution stored in dp[0][(1 << N) - 1].

Implementation:

C++

#include <bits/stdc++.h> using namespace std;  // Function to obtain maximum sum // of Bitwise AND of same-indexed // elements from the arrays A[] and B[] int maximizeAndUtil(int* A, int* B, int N) {     // dp[i][mask] stores the maximum sum of Bitwise AND     // of first i elements of A and jth element of B,     // where each element of B can be used at most once and     // the set of already used elements is represented by     // the binary mask 'mask'.     vector<vector<int> > dp(N + 1, vector<int>(1 << N, 0));      // dp[N][mask] is initialized to 0 for all masks     // as Bitwise AND of empty sets is 0.     for (int mask = 0; mask < (1 << N); mask++) {         dp[N][mask] = 0;     }      // i ranges from N-1 to 0     for (int i = N - 1; i >= 0; i--) {          // j ranges from 0 to N-1         for (int j = 0; j < N; j++) {              // If the j-th element of B is             // not already used             if ((1 << j) & ((1 << N) - 1)) {                  // Iterate over all possible                 // sets of elements of B                 // that can be used along                 // with the j-th element                 for (int mask = 0; mask < (1 << N);                      mask++) {                     if ((1 << j) & mask) {                         dp[i][mask] = max(                             dp[i][mask],                             (A[i] & B[j])                                 + dp[i + 1]                                     [mask ^ (1 << j)]);                     }                 }             }         }     }      // The maximum sum is stored     // in dp[0][(1 << N) - 1]     return dp[0][(1 << N) - 1]; }  // Driver Code int main() {     int A[] = { 3, 5, 7, 11 };     int B[] = { 2, 6, 10, 12 };     int N = sizeof A / sizeof A[0];      cout << maximizeAndUtil(A, B, N); }

Java

import java.util.*;  // Added by ~Nikunj Sonigara public class Main {     public static int maximizeAndUtil(int[] A, int[] B, int N) {         int[][] dp = new int[N + 1][1 << N];          for (int mask = 0; mask < (1 << N); mask++) {             dp[N][mask] = 0;         }          for (int i = N - 1; i >= 0; i--) {             for (int j = 0; j < N; j++) {                 if ((1 << j & (1 << N) - 1) != 0) {                     for (int mask = 0; mask < (1 << N); mask++) {                         if ((1 << j & mask) != 0) {                             dp[i][mask] = Math.max(dp[i][mask], (A[i] & B[j]) + dp[i + 1][mask ^ (1 << j)]);                         }                     }                 }             }         }          return dp[0][(1 << N) - 1];     }      public static void main(String[] args) {         int[] A = {3, 5, 7, 11};         int[] B = {2, 6, 10, 12};         int N = A.length;          System.out.println(maximizeAndUtil(A, B, N));     } }

Python3

# Added by ~Nikunj Sonigara  def maximizeAndUtil(A, B, N):     dp = [[0] * (1 << N) for _ in range(N + 1)]      for mask in range(1 << N):         dp[N][mask] = 0      for i in range(N - 1, -1, -1):         for j in range(N):             if (1 << j) & ((1 << N) - 1):                 for mask in range(1 << N):                     if (1 << j) & mask:                         dp[i][mask] = max(dp[i][mask], (A[i] & B[j]) + dp[i + 1][mask ^ (1 << j)])      return dp[0][(1 << N) - 1]  if __name__ == "__main__":     A = [3, 5, 7, 11]     B = [2, 6, 10, 12]     N = len(A)      print(maximizeAndUtil(A, B, N))

using System;  public class MaximizeAndUtilMain {     public static int MaximizeAndUtil(int[] A, int[] B, int N)     {         // Initialize a 2D array to store the dynamic programming results.         int[,] dp = new int[N + 1, 1 << N];          // Initialize the last row of the dynamic programming table to 0.         for (int mask = 0; mask < (1 << N); mask++)         {             dp[N, mask] = 0;         }          // Start filling the dynamic programming table from the second-to-last row, going backwards.         for (int i = N - 1; i >= 0; i--)         {             // Iterate through all elements in array B.             for (int j = 0; j < N; j++)             {                 // Check if the j-th element in array B is included in the current mask.                 if ((1 << j & ((1 << N) - 1)) != 0)                 {                     // Iterate through all possible masks.                     for (int mask = 0; mask < (1 << N); mask++)                     {                         // Check if the j-th element is included in the current mask.                         if ((1 << j & mask) != 0)                         {                             // Calculate the maximum of the current dp value and the bitwise AND of A[i] and B[j]                             // plus the value from the next row and the mask with j-th bit turned off.                             dp[i, mask] = Math.Max(dp[i, mask], (A[i] & B[j]) + dp[i + 1, mask ^ (1 << j)]);                         }                     }                 }             }         }          // The result is stored in the top-left cell of the dynamic programming table.         return dp[0, (1 << N) - 1];     }      public static void Main()     {         int[] A = { 3, 5, 7, 11 };         int[] B = { 2, 6, 10, 12 };         int N = A.Length;          // Call the MaximizeAndUtil function and print the result.         Console.WriteLine(MaximizeAndUtil(A, B, N));     } }

JavaScript

function maximizeAndUtil(A, B, N) {   // dp[i][mask] stores the maximum sum of Bitwise AND   // of first i elements of A and jth element of B,   // where each element of B can be used at most once and   // the set of already used elements is represented by   // the binary mask 'mask'.   const dp = new Array(N + 1).fill(0).map(() => new Array(1 << N).fill(0));    // dp[N][mask] is initialized to 0 for all masks   // as Bitwise AND of empty sets is 0.   for (let mask = 0; mask < (1 << N); mask++) {     dp[N][mask] = 0;   }    // i ranges from N-1 to 0   for (let i = N - 1; i >= 0; i--) {     // j ranges from 0 to N-1     for (let j = 0; j < N; j++) {       // If the j-th element of B is       // not already used       if ((1 << j) & ((1 << N) - 1)) {         // Iterate over all possible         // sets of elements of B         // that can be used along         // with the j-th element         for (let mask = 0; mask < (1 << N); mask++) {           if ((1 << j) & mask) {             dp[i][mask] = Math.max(               dp[i][mask],               (A[i] & B[j]) + dp[i + 1][mask ^ (1 << j)]             );           }         }       }     }   }    // The maximum sum is stored   // in dp[0][(1 << N) - 1]   return dp[0][(1 << N) - 1]; }  // Driver code const A = [3, 5, 7, 11]; const B = [2, 6, 10, 12]; const N = A.length;  console.log(maximizeAndUtil(A, B, N));

Output

Time Complexity: O (N² * 2^N)
Auxiliary Space: O(N * 2^N)

Maximize Array sum by adding multiple of another Array element in given ranges

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Rearrange an array to maximize sum of Bitwise AND of same-indexed elements with another array

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