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Find maximum product of Bitwise AND and Bitwise OR of K-size subarray

  • Difficulty Level : Expert
  • Last Updated : 22 Dec, 2021

Given an array arr[] containing N integers and an integer K, the task is to find the maximum value of the product of Bitwise AND and Bitwise OR of all elements of a K-sized subarray.

Example:

Input: arr[] = {1, 2, 3, 4}, K = 2
Output: 6
Explanation: Bitwise AND and Bitwise XOR of all K-sized subarrays is {(0, 3), (2, 3), (0, 7)} respectively. Therefore the maximum possible value of their product is 2 * 3 = 6 which is the required answer.

Input: arr[] = {6, 7, 7, 10, 8, 2}, K = 3
Output: 42

 

Approach: The given problem can be solved using the sliding window technique. The idea is to maintain the value of bitwise AND and bitwise OR for the K-sized subarray windows, which can be done by maintaining an array storing the count of each bit over all the elements of the current window. If the count of ith bit is K, that bit must be included in the bitwise AND and if the count of the bit is greater than 1, it must be included in the bitwise OR. Store the maximum value of their product over all the windows in a variable ans which is the required 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 convert bit array to
// decimal number representing OR
int build_or(int bit[])
{
    int ans = 0;
 
    for (int i = 0; i < 32; i++)
        if (bit[i])
            ans += (1 << i);
 
    return ans;
}
 
// Function to convert bit array to
// decimal number representing AND
int build_and(int bit[], int k)
{
    int ans = 0;
 
    for (int i = 0; i < 32; i++)
        if (bit[i] == k)
            ans += (1 << i);
 
    return ans;
}
 
// Function to find maximum value of
// AND * OR over all K sized subarrays
int maximizeAndOr(int arr[], int N, int K)
{
    // Maintain an integer array bit[]
    // of size 32 all initialized to 0
    int bit[32] = { 0 };
 
    // Create a sliding window of size k
    for (int i = 0; i < K; i++) {
        for (int j = 0; j < 32; j++) {
            if (arr[i] & (1 << j))
                bit[j]++;
        }
    }
 
    int ans = build_and(bit, K) * build_or(bit);
 
    for (int i = K; i < N; i++) {
 
        // Perform operation for
        // removed element
        for (int j = 0; j < 32; j++) {
            if (arr[i - K] & (1 << j))
                bit[j]--;
        }
 
        // Perform operation for
        // added_element
        for (int j = 0; j < 32; j++) {
            if (arr[i] & (1 << j))
                bit[j]++;
        }
 
        // Taking maximum value
        ans = max(ans, build_and(bit, K)
                  * build_or(bit));
    }
 
    return ans;
}
 
// Driver Code
int main()
{
    int arr[] = { 6, 7, 7, 10, 8, 2 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int K = 3;
 
    cout << maximizeAndOr(arr, N, K);
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
 
class GFG{
     
// Function to convert bit array to
// decimal number representing OR
static int build_or(int[] bit)
{
    int ans = 0;
 
    for(int i = 0; i < 32; i++)
        if (bit[i] > 0)
            ans += (1 << i);
             
    return ans;
}
 
// Function to convert bit array to
// decimal number representing AND
static int build_and(int[] bit, int k)
{
    int ans = 0;
 
    for(int i = 0; i < 32; i++)
        if (bit[i] == k)
            ans += (1 << i);
 
    return ans;
}
 
// Function to find maximum value of
// AND * OR over all K sized subarrays
static int maximizeAndOr(int[] arr, int N, int K)
{
     
    // Maintain an integer array bit[]
    // of size 32 all initialized to 0
    int[] bit = new int[32];
 
    // Create a sliding window of size k
    for(int i = 0; i < K; i++)
    {
        for(int j = 0; j < 32; j++)
        {
            if ((arr[i] & (1 << j)) > 0)
                bit[j]++;
        }
    }
 
    int ans = build_and(bit, K) * build_or(bit);
 
    for(int i = K; i < N; i++)
    {
         
        // Perform operation for
        // removed element
        for(int j = 0; j < 32; j++)
        {
            if ((arr[i - K] & (1 << j)) > 0)
                bit[j]--;
        }
 
        // Perform operation for
        // added_element
        for(int j = 0; j < 32; j++)
        {
            if ((arr[i] & (1 << j)) > 0)
                bit[j]++;
        }
 
        // Taking maximum value
        ans = Math.max(ans, build_and(bit, K) *
                            build_or(bit));
    }
    return ans;
}
 
// Driver Code
public static void main(String args[])
{
    int[] arr = { 6, 7, 7, 10, 8, 2 };
    int N = arr.length;
    int K = 3;
 
    System.out.println(maximizeAndOr(arr, N, K));
}
}
 
// This code is contributed by Samim Hossain Mondal.

Python3




# Python3 program for the above approach
 
# Function to convert bit array to
# decimal number representing OR
def build_or(bit):
     
    ans = 0
 
    for i in range(0, 32):
        if (bit[i]):
            ans += (1 << i)
 
    return ans
 
# Function to convert bit array to
# decimal number representing AND
def build_and(bit, k):
 
    ans = 0
 
    for i in range(0, 32):
        if (bit[i] == k):
            ans += (1 << i)
 
    return ans
 
# Function to find maximum value of
# AND * OR over all K sized subarrays
def maximizeAndOr(arr, N, K):
 
    # Maintain an integer array bit[]
    # of size 32 all initialized to 0
    bit = [0 for _ in range(32)]
 
    # Create a sliding window of size k
    for i in range(0, K):
        for j in range(0, 32):
            if (arr[i] & (1 << j)):
                bit[j] += 1
 
    ans = build_and(bit, K) * build_or(bit)
 
    for i in range(K, N):
 
        # Perform operation for
        # removed element
        for j in range(0, 32):
            if (arr[i - K] & (1 << j)):
                bit[j] -= 1
 
        # Perform operation for
        # added_element
        for j in range(0, 32):
            if (arr[i] & (1 << j)):
                bit[j] += 1
 
        # Taking maximum value
        ans = max(ans, build_and(bit, K) *
                       build_or(bit))
 
    return ans
 
# Driver Code
if __name__ == "__main__":
 
    arr = [ 6, 7, 7, 10, 8, 2 ]
    N = len(arr)
    K = 3
 
    print(maximizeAndOr(arr, N, K))
 
# This code is contributed by rakeshsahni

C#




// C# program for the above approach
using System;
class GFG
{
   
    // Function to convert bit array to
    // decimal number representing OR
    static int build_or(int[] bit)
    {
        int ans = 0;
 
        for (int i = 0; i < 32; i++)
            if (bit[i] > 0)
                ans += (1 << i);
 
        return ans;
    }
 
    // Function to convert bit array to
    // decimal number representing AND
    static int build_and(int[] bit, int k)
    {
        int ans = 0;
 
        for (int i = 0; i < 32; i++)
            if (bit[i] == k)
                ans += (1 << i);
 
        return ans;
    }
 
    // Function to find maximum value of
    // AND * OR over all K sized subarrays
    static int maximizeAndOr(int[] arr, int N, int K)
    {
        // Maintain an integer array bit[]
        // of size 32 all initialized to 0
        int[] bit = new int[32];
 
        // Create a sliding window of size k
        for (int i = 0; i < K; i++) {
            for (int j = 0; j < 32; j++) {
                if ((arr[i] & (1 << j)) > 0)
                    bit[j]++;
            }
        }
 
        int ans = build_and(bit, K) * build_or(bit);
 
        for (int i = K; i < N; i++) {
 
            // Perform operation for
            // removed element
            for (int j = 0; j < 32; j++) {
                if ((arr[i - K] & (1 << j)) > 0)
                    bit[j]--;
            }
 
            // Perform operation for
            // added_element
            for (int j = 0; j < 32; j++) {
                if ((arr[i] & (1 << j)) > 0)
                    bit[j]++;
            }
 
            // Taking maximum value
            ans = Math.Max(ans, build_and(bit, K)
                                    * build_or(bit));
        }
 
        return ans;
    }
 
    // Driver Code
    public static void Main()
    {
        int[] arr = { 6, 7, 7, 10, 8, 2 };
        int N = arr.Length;
        int K = 3;
 
        Console.WriteLine(maximizeAndOr(arr, N, K));
    }
}
 
// This code is contributed by ukasp.

Javascript




<script>
 
// JavaScript program to implement
// the above approach
 
// Function to convert bit array to
// decimal number representing OR
function build_or(bit)
{
    let ans = 0;
 
    for(let i = 0; i < 32; i++)
        if (bit[i])
            ans += (1 << i);
 
    return ans;
}
 
// Function to convert bit array to
// decimal number representing AND
function build_and(bit, k)
{
    let ans = 0;
 
    for(let i = 0; i < 32; i++)
        if (bit[i] == k)
            ans += (1 << i);
 
    return ans;
}
 
// Function to find maximum value of
// AND * OR over all K sized subarrays
function maximizeAndOr(arr, N, K)
{
     
    // Maintain an integer array bit[]
    // of size 32 all initialized to 0
    let bit = new Array(32).fill(0);
 
    // Create a sliding window of size k
    for(let i = 0; i < K; i++)
    {
        for(let j = 0; j < 32; j++)
        {
            if (arr[i] & (1 << j))
                bit[j]++;
        }
    }
 
    let ans = build_and(bit, K) * build_or(bit);
 
    for(let i = K; i < N; i++)
    {
         
        // Perform operation for
        // removed element
        for(let j = 0; j < 32; j++)
        {
            if (arr[i - K] & (1 << j))
                bit[j]--;
        }
 
        // Perform operation for
        // added_element
        for(let j = 0; j < 32; j++)
        {
            if (arr[i] & (1 << j))
                bit[j]++;
        }
 
        // Taking maximum value
        ans = Math.max(ans, build_and(bit, K) *
                            build_or(bit));
    }
    return ans;
}
 
// Driver Code
let arr = [ 6, 7, 7, 10, 8, 2 ];
let N = arr.length;
let K = 3;
 
document.write(maximizeAndOr(arr, N, K));
 
// This code is contributed by Potta Lokesh
 
</script>
Output
42

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

 


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