Rearrange an array to maximize sum of Bitwise AND of same-indexed elements with another array
Last Updated :
23 Jul, 2025
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 jth 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*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.
- 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 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).
- 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 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 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.
<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>
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 <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);
}
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));
}
}
# 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));
}
}
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));
Time Complexity: O (N2 * 2N)
Auxiliary Space: O(N * 2N)
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