Bitwise OR of bitwise AND of all possible non-empty subarrays after Q query updates

Given an array arr[] consisting of N positive integers and an array of queries Q[] of type [L, R], the task is to find the Bitwise OR of the bitwise AND of all the possible non-empty subarrays of the array after updating array element at index L to R for each query.

Examples:

Input: arr[ ] = {1, 2, 3}, Q[ ] = {{1, 4}, {3, 0}}
Output: 7 6
Explanation:
All subarrays are {1}, {2}, {3}, {1, 2}, {2, 3} and {1, 2, 3}
For query 1: Update arr[1] = 4, then new array is {4, 2, 3}. The bitwise & of all subarrays are 4, 2, 3, 0, 2, 0 and the bitwise OR( {4, 2, 3, 0, 2, 0}) equals 7.
For query 2: Update arr[3] = 0, then new array is {4, 2, 0}. The bitwise & of all subarrays are 4, 2, 0, 0, 0, 0 and the bitwise OR( {4, 2, 0, 0, 0, 0}) equals 6.

Input: arr[ ] = {1, 2, 1}, Q[ ] = {{2, 1}}
Output: 1

Naive Approach: The simplest approach to solve the problem is to traverse the array Q[] and for each query update the array element arr[L] to R and find all the subarrays and their bitwise AND and store them in the new array. After storing the Bitwise AND, find the bitwise OR of the new array formed.

Time Complexity: O(N²Q)
Auxiliary Space: O(N²)

Efficient Approach: The above approach can also be optimized by using the fact that the bitwise OR of the resulting bitwise AND of all the generated subarrays are the same as the resulting bitwise OR of all the elements present in the array.

Therefore, the idea is to perform the queries and print the value of Bitwise OR of all array elements after updating the array each time.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include<bits/stdc++.h>
using namespace std;


    // Function to find the Bitwise OR
    // of Bitwise AND of all possible
    // subarrays after performing the
    // every query
  void performQuery(vector<int> arr,
                             vector<vector<int>> Q)
    {
        // Traversing each pair
        // of the query
        for (int i = 0; i < Q.size(); i++) {

            // Stores the Bitwise OR
            int or1 = 0;

            // Updating the array
            int x = Q[i][0];
            arr[x - 1] = Q[i][1];

            // Find the Bitwise OR of
            // new updated array
            for (int j = 0; j < arr.size(); j++) {
                or1 = or1 | arr[j];
            }

            // Print the ans
            cout<<or1<<" ";
        }
    }

    // Driver Code
   int main()
    {
        vector<int> arr({ 1, 2, 3 });
        vector<int> v1({1,4});
        vector<int> v2({3,0});
        vector<vector<int>> Q;
        Q.push_back(v1);
        Q.push_back(v2);
        performQuery(arr, Q);
    }

// This code is contributed by ipg2016107.

Java

// Java program for the above approach
import java.io.*;

class GFG {

    // Function to find the Bitwise OR
    // of Bitwise AND of all possible
    // subarrays after performing the
    // every query
    static void performQuery(int arr[],
                             int Q[][])
    {
        // Traversing each pair
        // of the query
        for (int i = 0; i < Q.length; i++) {

            // Stores the Bitwise OR
            int or = 0;

            // Updating the array
            int x = Q[i][0];
            arr[x - 1] = Q[i][1];

            // Find the Bitwise OR of
            // new updated array
            for (int j = 0; j < arr.length; j++) {
                or = or | arr[j];
            }

            // Print the ans
            System.out.print(or + " ");
        }
    }

    // Driver Code
    public static void main(String[] args)
    {
        int arr[] = { 1, 2, 3 };
        int Q[][] = { { 1, 4 }, { 3, 0 } };
        performQuery(arr, Q);
    }
}

Python3

# Python program for the above approach
# Function to find the Bitwise OR
# of Bitwise AND of all possible
# subarrays after performing the
# every query
def performQuery(arr, Q):

    # Traversing each pair
    # of the query
    for i in range (0, len(Q)):

        # Stores the Bitwise OR
        orr = 0

        # Updating the array
        x = Q[i][0]
        arr[x - 1] = Q[i][1]

        # Find the Bitwise OR of
        # new updated array
        for j in range(0,len(arr)):
            orr = orr | arr[j]

        # Print the ans
        print(orr ,end= " ")

# Driver Code
arr = [1, 2, 3]
Q = [[1, 4] , [3, 0]]
performQuery(arr, Q)

# This code is contributed by shivanisinghss2110

C#

// C# program for the above approach
using System;

class GFG {

    // Function to find the Bitwise OR
    // of Bitwise AND of all possible
    // subarrays after performing the
    // every query
    static void performQuery(int []arr, int [,]Q)
    {
        // Traversing each pair
        // of the query
        for (int i = 0; i < Q.Length; i++) {

            // Stores the Bitwise OR
            int or = 0;

            // Updating the array
            int x = Q[i,0];
            arr[x - 1] = Q[i,1];

            // Find the Bitwise OR of
            // new updated array
            for (int j = 0; j < arr.Length; j++) {
                or = or | arr[j];
            }

            // Print the ans
            Console.Write(or + " ");
        }
    }

    // Driver Code
    public static void Main(String[] args)
    {
        int []arr = { 1, 2, 3 };
        int [,]Q = { { 1, 4 }, { 3, 0 } };
        performQuery(arr, Q);
    }
}

// This code is contributed by shivanisinghss2110

Javascript

<script>
        // JavaScript Program to implement
        // the above approach
        // Function to find the Bitwise OR
        // of Bitwise AND of all possible
        // subarrays after performing the
        // every query
        function performQuery(arr, Q)
        {

            // Traversing each pair
            // of the query
            for (let i = 0; i < Q.length; i++) {

                // Stores the Bitwise OR
                let or = 0;

                // Updating the array
                let x = Q[i][0];
                arr[x - 1] = Q[i][1];

                // Find the Bitwise OR of
                // new updated array
                for (let j = 0; j < arr.length; j++) {
                    or = or | arr[j];
                }

                // Print the ans
                document.write(or + " ");
            }
        }

        // Driver Code
        let arr = [1, 2, 3];
        let Q = [[1, 4], [3, 0]];
        performQuery(arr, Q);

// This code is contributed by Potta Lokesh
    </script>

Output:

7 6

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