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

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]:

Follow the steps below to solve the problem:

• Define a vector of vectors, says dp of dimension N*2N  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.
• Iterate over the range [0, N-1] using variable j and in each iteration, If jth bit in the mask is not set then update dp[i][mask] as dp[i][mask] = max(dp[i][mask], maximizeUtil(i+1, mask| 2j).
• 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 ` `using` `namespace` `std;`   `// Function to implement recursive DP` `int` `maximizeAnd(``int` `i, ``int` `mask,` `                ``int``* A, ``int``* B, ``int` `N,` `                ``vector >& 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 > 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#

 `// 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

## Javascript

 ``

Output

```22

```

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

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 jth 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 ` `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 > 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))`

## C#

 `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

```22

```

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

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