# Bitwise OR of Bitwise AND of all subsets of an Array for Q queries

Given two arrays arr[] of size N and queries[] of size Q, the task is to find the OR of AND of subsets of the array. In each query, you are given an index and a value, you have to replace the value at the given index of the arrays with a given value and print the OR of AND of all subsets of the array after each query.

Examples:

Input: arr[] = {3, 5, 7}, N = 3, queries[] = {{1, 2}, {2, 1}}, Q = 2
Output:
3
Explanation:
For the first query replace the value at index 1 with value 2, updated array arr[] is {3, 2, 7}.
Then take AND of all subsets i.e. AND(3) = 3, AND(2) = 2, AND(7) = 7, AND(3, 2) = 2, AND(3, 7) = 3, AND(3, 2, 7) = 2.
OR of the AND of all subsets are OR(3, 2, 7, 2, 3, 2) = 7.
Now, for the second query replace the value at index 2 with value 1, updated array arr[] is {3, 2, 1}.
Then take AND of all subsets i.e. AND(3) = 3, AND(2) = 2, AND(1) = 1, AND(3, 2) = 2, AND(3, 1) = 1, AND(3, 2, 1) = 0.
OR of the AND of all subsets OR(3, 2, 1, 2, 1, 0) = 3.

Input: arr[] = {1, 2, 3}, N = 3, queries[] = {{2, 4}, {1, 8}}, Q = 2
Output: 7
13

Approach: This problem can be solved using greedy algorithm. Follow the steps below to solve the problem:

• Initialize an array bits[] of size 32 and store the count of set bits of all the elements.
• Iterate in the range [0, Q-1] using the variable p:
• First subtract the bits of previous value and then add the bits of new value.
• Iterate in the range [0, 31] using the variable i:
• If the current bit is set for the previous value, then subtract one bit to the bits[] array at ith index.
• If the current bit is set for the new value, then add one bit to the bits[] array at ith index.
• Replace the new value from the previous value in given array arr[].
• Initialize a variable ans to store the OR of the bits array.
• Iterate in the range [0, 31] using the variable i:
• If the current bit is not equal to 0, then add the OR of the current bit into ans.
• After completing the above steps, print the ans as the required answer for each query.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to find the OR of AND of all``// subsets of the array for each query``void` `Or_of_Ands_for_each_query(``int` `arr[], ``int` `n,``                               ``int` `queries[][2], ``int` `q)``{``    ``// An array to store the bits``    ``int` `bits[32] = { 0 };` `    ``// Iterate for all the bits``    ``for` `(``int` `i = 0; i < 32; i++) {``        ``// Iterate over all the numbers and``        ``// store the bits in bits[] array``        ``for` `(``int` `j = 0; j < n; j++) {``            ``if` `((1 << i) & arr[j]) {``                ``bits[i]++;``            ``}``        ``}``    ``}` `    ``// Iterate over all the queries``    ``for` `(``int` `p = 0; p < q; p++) {` `        ``// Replace the bits of the value``        ``// at arr[queries[p][0]] with the bits``        ``// of queries[p][1]``        ``for` `(``int` `i = 0; i < 32; i++) {``            ``if` `((1 << i) & arr[queries[p][0]]) {``                ``bits[i]--;``            ``}``            ``if` `(queries[p][1] & (1 << i)) {``                ``bits[i]++;``            ``}``        ``}` `        ``// Substitute the value in the array``        ``arr[queries[p][0]] = queries[p][1];``        ``int` `ans = 0;` `        ``// Find OR of the bits[] array``        ``for` `(``int` `i = 0; i < 32; i++) {``            ``if` `(bits[i] != 0) {``                ``ans |= (1 << i);``            ``}``        ``}``        ``// Print the answer``        ``cout << ans << endl;``    ``}``}` `// Driver Code``int` `main()``{``    ``// Given Input``    ``int` `n = 3, q = 2;``    ``int` `arr[] = { 3, 5, 7 };``    ``int` `queries[2][2] = { { 1, 2 }, { 2, 1 } };` `    ``// Function Call``    ``Or_of_Ands_for_each_query(arr, n, queries, q);` `    ``return` `0;``}`

## Java

 `// java program for the above approach``import` `java.util.*;` `class` `GFG {` `    ``// Function to find the OR of AND of all``    ``// subsets of the array for each query``    ``static` `void` `Or_of_Ands_for_each_query(``int` `arr[], ``int` `n,``                                          ``int` `queries[][],``                                          ``int` `q)``    ``{``        ``// An array to store the bits``        ``int` `bits[] = ``new` `int``[``32``];``        ``Arrays.fill(bits, ``0``);` `        ``// Iterate for all the bits``        ``for` `(``int` `i = ``0``; i < ``32``; i++) {``            ``// Iterate over all the numbers and``            ``// store the bits in bits[] array``            ``for` `(``int` `j = ``0``; j < n; j++) {``                ``if` `(((``1` `<< i) & arr[j]) != ``0``) {``                    ``bits[i]++;``                ``}``            ``}``        ``}` `        ``// Iterate over all the queries``        ``for` `(``int` `p = ``0``; p < q; p++) {` `            ``// Replace the bits of the value``            ``// at arr[queries[p][0]] with the bits``            ``// of queries[p][1]``            ``for` `(``int` `i = ``0``; i < ``32``; i++) {``                ``if` `(((``1` `<< i) & arr[queries[p][``0``]]) != ``0``) {``                    ``bits[i]--;``                ``}``                ``if` `((queries[p][``1``] & (``1` `<< i)) != ``0``) {``                    ``bits[i]++;``                ``}``            ``}` `            ``// Substitute the value in the array``            ``arr[queries[p][``0``]] = queries[p][``1``];``            ``int` `ans = ``0``;` `            ``// Find OR of the bits[] array``            ``for` `(``int` `i = ``0``; i < ``32``; i++) {``                ``if` `(bits[i] != ``0``) {``                    ``ans |= (``1` `<< i);``                ``}``            ``}` `            ``// Print the answer``            ``System.out.println(ans);``        ``}``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{` `        ``// Given Input``        ``int` `n = ``3``, q = ``2``;``        ``int` `arr[] = { ``3``, ``5``, ``7` `};``        ``int` `queries[][] = { { ``1``, ``2` `}, { ``2``, ``1` `} };` `        ``// Function Call``        ``Or_of_Ands_for_each_query(arr, n, queries, q);``    ``}``}` `// This code is contributed by subhammahato348.`

## Python3

 `# Python3 program for the above approach` `# Function to find the OR of AND of all``# subsets of the array for each query`  `def` `Or_of_Ands_for_each_query(arr, n, queries, q):` `    ``# An array to store the bits``    ``bits ``=` `[``0` `for` `x ``in` `range``(``32``)]` `    ``# Iterate for all the bits``    ``for` `i ``in` `range``(``0``, ``32``):` `        ``# Iterate over all the numbers and``        ``# store the bits in bits[] array``        ``for` `j ``in` `range``(``0``, n):``            ``if` `((``1` `<< i) & arr[j]):``                ``bits[i] ``+``=` `1` `    ``# Iterate over all the queries``    ``for` `p ``in` `range``(``0``, q):` `        ``# Replace the bits of the value``        ``# at arr[queries[p][0]] with the bits``        ``# of queries[p][1]``        ``for` `i ``in` `range``(``0``, ``32``):``            ``if` `((``1` `<< i) & arr[queries[p][``0``]]):``                ``bits[i] ``-``=` `1``            ``if` `(queries[p][``1``] & (``1` `<< i)):``                ``bits[i] ``+``=` `1` `        ``# Substitute the value in the array``        ``arr[queries[p][``0``]] ``=` `queries[p][``1``]``        ``ans ``=` `0` `        ``# Find OR of the bits[] array``        ``for` `i ``in` `range``(``0``, ``32``):``            ``if` `(bits[i] !``=` `0``):``                ``ans |``=` `(``1` `<< i)` `        ``# Print the answer``        ``print``(ans)` `#  Driver Code`  `#  Given Input``n ``=` `3``q ``=` `2``arr ``=` `[``3``, ``5``, ``7``]``queries ``=` `[[``1``, ``2``], [``2``, ``1``]]` `# Function Call``Or_of_Ands_for_each_query(arr, n, queries, q)` `# This code is contributed by amreshkumar3`

## C#

 `// C# program for the above approach``using` `System;` `class` `GFG{` `// Function to find the OR of AND of all``// subsets of the array for each query``static` `void` `Or_of_Ands_for_each_query(``int``[] arr, ``int` `n,``                                      ``int``[,] queries,``                                      ``int` `q)``{``    ` `    ``// An array to store the bits``    ``int``[] bits = ``new` `int``[32];``    ``for``(``int` `i = 0; i < 32; i++) ``    ``{``        ``bits[i] = 0;``    ``}` `    ``// Iterate for all the bits``    ``for``(``int` `i = 0; i < 32; i++) ``    ``{``        ` `        ``// Iterate over all the numbers and``        ``// store the bits in bits[] array``        ``for``(``int` `j = 0; j < n; j++)``        ``{``            ``if` `(((1 << i) & arr[j]) != 0) ``            ``{``                ``bits[i]++;``            ``}``        ``}``    ``}` `    ``// Iterate over all the queries``    ``for``(``int` `p = 0; p < q; p++)``    ``{``        ` `        ``// Replace the bits of the value``        ``// at arr[queries[p][0]] with the bits``        ``// of queries[p][1]``        ``for``(``int` `i = 0; i < 32; i++) ``        ``{``            ``if` `(((1 << i) & arr[queries[p, 0]]) != 0)``            ``{``                ``bits[i]--;``            ``}``            ``if` `((queries[p, 1] & (1 << i)) != 0)``            ``{``                ``bits[i]++;``            ``}``        ``}` `        ``// Substitute the value in the array``        ``arr[queries[p, 0]] = queries[p, 1];``        ``int` `ans = 0;` `        ``// Find OR of the bits[] array``        ``for``(``int` `i = 0; i < 32; i++)``        ``{``            ``if` `(bits[i] != 0)``            ``{``                ``ans |= (1 << i);``            ``}``        ``}` `        ``// Print the answer``        ``Console.WriteLine(ans);``    ``}``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``    ` `    ``// Given Input``    ``int` `n = 3, q = 2;``    ``int``[] arr = { 3, 5, 7 };``    ``int``[,] queries = { { 1, 2 }, { 2, 1 } };``    ` `    ``// Function Call``    ``Or_of_Ands_for_each_query(arr, n, queries, q);``}``}` `// This code is contributed by target_2`

## Javascript

 ``

Output
```7
3```

Time Complexity: O(max(N, Q))
Auxiliary Space: O(1)

Previous
Next