Queries to calculate Bitwise AND of an array with updates

Given an array arr[] consisting of N positive integers and a 2D array Q[][] consisting of queries of the type {i, val}, the task for each query is to replace arr[i] by val and calculate the Bitwise AND of the modified array.

Examples:

Input: arr[] = {1, 2, 3, 4, 5}, Q[][] = {{0, 2}, {3, 3}, {4, 2}}
Output: 0 0 2
Explanation:
Query 1: Update A[0] to 2, then the array modifies to {2, 2, 3, 4, 5}. The Bitwise AND of all the elements is 0.
Query 2: Update A[3] to 3, then the array modifies to {2, 2, 3, 3, 5}. The Bitwise AND of all the elements is 0.
Query 3: Update A[4] to 2, then the modified array, A[]={2, 2, 3, 3, 2}. The Bitwise AND of all the elements is 2.

Input: arr[] = {1, 2, 3}, Q[][] = {{1, 5}, {2, 4}}
Output: 1 0

Naive Approach: The simplest approach is to solve the given problem is to update the array element for each query and then find the bitwise AND of all the array elements by traversing the array in each query.

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

Efficient Approach: The above approach can also be optimized by using an auxiliary array, say bitCount[][] of size 32 * N to store the sum of set bits at position i till the j^th index of the array. So, bitCount[i][N – 1] represents the total sum of set bits at position i using all the array elements. Follow the steps below to solve the problem:

• Initialize an array bitCount[32][N] to store the set bits of the array elements.
• Iterate over the range [0, 31] using the variable i and perform the following steps:
  • If the value of A[0] is set at the i^th position, then update bitCount[i][0] to 1. Otherwise, update it to 0.
  • Traverse the array A[] in the range [1, N – 1] using the variable j and perform the following steps:
    • If the value of A[j] is set at i^th position, then update bitCount[i][j] to 1.
    • Add the value of bitCount[i][j – 1] to bitCount[i][j].
• Traverse the given array of queries Q[][] and perform the following steps:
  • Initialize a variable, say res as 0, to store the result of the current query.
  • Store the current value at the given index in currentVal and the new value in newVal.
  • Iterate over the range [0, 31] using the variable i
    • If newVal is set at index i and currentVal is not set, then increment prefix[i][N – 1] by 1.
    • Otherwise, if currentVal is set at index i and newVal is not set, then decrement prefix[i][N – 1] by 1.
    • If the value of prefix[i][N – 1] is equal to N, set this bit in res.
  • After completing the above steps, print the value of res as the result.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Store the number of set bits at
// each position
int prefixCount[32][10000];

// Function to precompute the prefix
// count array
void findPrefixCount(vector<int> arr,
                     int size)
{
    // Iterate over the range[0, 31]
    for (int i = 0; i < 32; i++) {

        // Set the bit at position i
        // if arr[0] is set at position i
        prefixCount[i][0]
            = ((arr[0] >> i) & 1);

        // Traverse the array and
        // take prefix sum
        for (int j = 1; j < size; j++) {

            // Update prefixCount[i][j]
            prefixCount[i][j]
                = ((arr[j] >> i) & 1);
            prefixCount[i][j]
                += prefixCount[i][j - 1];
        }
    }
}

// Function to find the Bitwise AND
// of all array elements
void arrayBitwiseAND(int size)
{
    // Stores the required result
    int result = 0;

    // Iterate over the range [0, 31]
    for (int i = 0; i < 32; i++) {

        // Stores the number of set
        // bits at position i
        int temp = prefixCount[i]
                              [size - 1];

        // If temp is N, then set ith
        // position in the result
        if (temp == size)
            result = (result | (1 << i));
    }

    // Print the result
    cout << result << " ";
}

// Function to update the prefix count
// array in each query
void applyQuery(int currentVal, int newVal,
                int size)
{
    // Iterate through all the bits
    // of the current number
    for (int i = 0; i < 32; i++) {

        // Store the bit at position
        // i in the current value and
        // the new value
        int bit1 = ((currentVal >> i) & 1);
        int bit2 = ((newVal >> i) & 1);

        // If bit2 is set and bit1 is
        // unset, then increase the
        // set bits at position i by 1
        if (bit2 > 0 && bit1 == 0)
            prefixCount[i][size - 1]++;

        // If bit1 is set and bit2 is
        // unset, then decrease the
        // set bits at position i by 1
        else if (bit1 > 0 && bit2 == 0)
            prefixCount[i][size - 1]--;
    }
}

// Function to find the bitwise AND
// of the array after each query
void findbitwiseAND(
    vector<vector<int> > queries,
    vector<int> arr, int N, int M)
{
    // Fill the prefix count array
    findPrefixCount(arr, N);

    // Traverse the queries
    for (int i = 0; i < M; i++) {

        // Store the index and
        // the new value
        int id = queries[i][0];
        int newVal = queries[i][1];

        // Store the current element
        // at the index
        int currentVal = arr[id];

        // Update the array element
        arr[id] = newVal;

        // Apply the changes to the
        // prefix count array
        applyQuery(currentVal, newVal, N);

        // Print the bitwise AND of
        // the modified array
        arrayBitwiseAND(N);
    }
}

// Driver Code
int main()
{
    vector<int> arr{ 1, 2, 3, 4, 5 };
    vector<vector<int> > queries{ { 0, 2 },
                                  { 3, 3 },
                                  { 4, 2 } };
    int N = arr.size();
    int M = queries.size();
    findbitwiseAND(queries, arr, N, M);

    return 0;
}

Java

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

class GFG{

// Store the number of set bits at
// each position
static int prefixCount[][];

// Function to precompute the prefix
// count array
static void findPrefixCount(int arr[], int size)
{

    // Iterate over the range[0, 31]
    for(int i = 0; i < 32; i++)
    {

        // Set the bit at position i
        // if arr[0] is set at position i
        prefixCount[i][0] = ((arr[0] >> i) & 1);

        // Traverse the array and
        // take prefix sum
        for(int j = 1; j < size; j++)
        {

            // Update prefixCount[i][j]
            prefixCount[i][j] = ((arr[j] >> i) & 1);
            prefixCount[i][j] += prefixCount[i][j - 1];
        }
    }
}

// Function to find the Bitwise AND
// of all array elements
static void arrayBitwiseAND(int size)
{

    // Stores the required result
    int result = 0;

    // Iterate over the range [0, 31]
    for(int i = 0; i < 32; i++)
    {

        // Stores the number of set
        // bits at position i
        int temp = prefixCount[i][size - 1];

        // If temp is N, then set ith
        // position in the result
        if (temp == size)
            result = (result | (1 << i));
    }

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

// Function to update the prefix count
// array in each query
static void applyQuery(int currentVal, int newVal,
                       int size)
{

    // Iterate through all the bits
    // of the current number
    for(int i = 0; i < 32; i++)
    {

        // Store the bit at position
        // i in the current value and
        // the new value
        int bit1 = ((currentVal >> i) & 1);
        int bit2 = ((newVal >> i) & 1);

        // If bit2 is set and bit1 is
        // unset, then increase the
        // set bits at position i by 1
        if (bit2 > 0 && bit1 == 0)
            prefixCount[i][size - 1]++;

        // If bit1 is set and bit2 is
        // unset, then decrease the
        // set bits at position i by 1
        else if (bit1 > 0 && bit2 == 0)
            prefixCount[i][size - 1]--;
    }
}

// Function to find the bitwise AND
// of the array after each query
static void findbitwiseAND(int queries[][], int arr[],
                           int N, int M)
{

    prefixCount = new int[32][10000];

    // Fill the prefix count array
    findPrefixCount(arr, N);

    // Traverse the queries
    for(int i = 0; i < M; i++)
    {

        // Store the index and
        // the new value
        int id = queries[i][0];
        int newVal = queries[i][1];

        // Store the current element
        // at the index
        int currentVal = arr[id];

        // Update the array element
        arr[id] = newVal;

        // Apply the changes to the
        // prefix count array
        applyQuery(currentVal, newVal, N);

        // Print the bitwise AND of
        // the modified array
        arrayBitwiseAND(N);
    }
}

// Driver Code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 3, 4, 5 };
    int queries[][] = { { 0, 2 }, { 3, 3 },
                        { 4, 2 } };
    int N = arr.length;
    int M = queries.length;

    findbitwiseAND(queries, arr, N, M);
}
}

// This code is contributed by Kingash

Python3

# Python3 program for the above approach

# Store the number of set bits at
# each position
prefixCount = [[0 for x in range(32)]
                  for y in range(10000)]

# Function to precompute the prefix
# count array
def findPrefixCount(arr, size):

    # Iterate over the range[0, 31]
    for i in range(32):

        # Set the bit at position i
        # if arr[0] is set at position i
        prefixCount[i][0] = ((arr[0] >> i) & 1)

        # Traverse the array and
        # take prefix sum
        for j in range(1, size):

            # Update prefixCount[i][j]
            prefixCount[i][j] = ((arr[j] >> i) & 1)
            prefixCount[i][j] += prefixCount[i][j - 1]

# Function to find the Bitwise AND
# of all array elements
def arrayBitwiseAND(size):

    # Stores the required result
    result = 0

    # Iterate over the range [0, 31]
    for i in range(32):

        # Stores the number of set
        # bits at position i
        temp = prefixCount[i][size - 1]

        # If temp is N, then set ith
        # position in the result
        if (temp == size):
            result = (result | (1 << i))

    # Print the result
    print(result, end = " ")

# Function to update the prefix count
# array in each query
def applyQuery(currentVal, newVal, size):

    # Iterate through all the bits
    # of the current number
    for i in range(32):

        # Store the bit at position
        # i in the current value and
        # the new value
        bit1 = ((currentVal >> i) & 1)
        bit2 = ((newVal >> i) & 1)

        # If bit2 is set and bit1 is
        # unset, then increase the
        # set bits at position i by 1
        if (bit2 > 0 and bit1 == 0):
            prefixCount[i][size - 1] += 1

        # If bit1 is set and bit2 is
        # unset, then decrease the
        # set bits at position i by 1
        elif (bit1 > 0 and bit2 == 0):
            prefixCount[i][size - 1] -= 1

# Function to find the bitwise AND
# of the array after each query
def findbitwiseAND(queries, arr, N,  M):

    # Fill the prefix count array
    findPrefixCount(arr, N)

    # Traverse the queries
    for i in range(M):

        # Store the index and
        # the new value
        id = queries[i][0]
        newVal = queries[i][1]

        # Store the current element
        # at the index
        currentVal = arr[id]

        # Update the array element
        arr[id] = newVal

        # Apply the changes to the
        # prefix count array
        applyQuery(currentVal, newVal, N)

        # Print the bitwise AND of
        # the modified array
        arrayBitwiseAND(N)

# Driver Code
if __name__ == "__main__":

    arr = [ 1, 2, 3, 4, 5 ]
    queries = [ [ 0, 2 ],
                [ 3, 3 ],
                [ 4, 2 ] ]
    N = len(arr)
    M = len(queries)

    findbitwiseAND(queries, arr, N, M)

# This code is contributed by ukasp

C#

// C# program for the above approach
using System;
using System.Collections.Generic;

class GFG{

// Store the number of set bits at
// each position
static int [,]prefixCount = new int[32, 10000];

// Function to precompute the prefix
// count array
static void findPrefixCount(List<int> arr,
                            int size)
{

    // Iterate over the range[0, 31]
    for(int i = 0; i < 32; i++)
    {

        // Set the bit at position i
        // if arr[0] is set at position i
        prefixCount[i, 0] = ((arr[0] >> i) & 1);

        // Traverse the array and
        // take prefix sum
        for(int j = 1; j < size; j++)
        {

            // Update prefixCount[i][j]
            prefixCount[i, j] = ((arr[j] >> i) & 1);
            prefixCount[i, j] += prefixCount[i, j - 1];
        }
    }
}

// Function to find the Bitwise AND
// of all array elements
static void arrayBitwiseAND(int size)
{

    // Stores the required result
    int result = 0;

    // Iterate over the range [0, 31]
    for(int i = 0; i < 32; i++)
    {

        // Stores the number of set
        // bits at position i
        int temp = prefixCount[i, size - 1];

        // If temp is N, then set ith
        // position in the result
        if (temp == size)
            result = (result | (1 << i));
    }

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

// Function to update the prefix count
// array in each query
static void applyQuery(int currentVal, int newVal,
                       int size)
{

    // Iterate through all the bits
    // of the current number
    for(int i = 0; i < 32; i++)
    {

        // Store the bit at position
        // i in the current value and
        // the new value
        int bit1 = ((currentVal >> i) & 1);
        int bit2 = ((newVal >> i) & 1);

        // If bit2 is set and bit1 is
        // unset, then increase the
        // set bits at position i by 1
        if (bit2 > 0 && bit1 == 0)
            prefixCount[i, size - 1]++;

        // If bit1 is set and bit2 is
        // unset, then decrease the
        // set bits at position i by 1
        else if (bit1 > 0 && bit2 == 0)
            prefixCount[i, size - 1]--;
    }
}

// Function to find the bitwise AND
// of the array after each query
static void findbitwiseAND(int [,]queries,
                      List<int> arr, int N, int M)
{

    // Fill the prefix count array
    findPrefixCount(arr, N);

    // Traverse the queries
    for(int i = 0; i < M; i++)
    {

        // Store the index and
        // the new value
        int id = queries[i,0];
        int newVal = queries[i,1];

        // Store the current element
        // at the index
        int currentVal = arr[id];

        // Update the array element
        arr[id] = newVal;

        // Apply the changes to the
        // prefix count array
        applyQuery(currentVal, newVal, N);

        // Print the bitwise AND of
        // the modified array
        arrayBitwiseAND(N);
    }
}

// Driver Code
public static void Main()
{
    List<int> arr = new List<int>(){ 1, 2, 3, 4, 5 };
    int [,] queries = new int [3, 2]{ { 0, 2 },
                                      { 3, 3 },
                                      { 4, 2 } };

    int N = arr.Count;
    int M = 3;

    findbitwiseAND(queries, arr, N, M);
}
}

// This code is contributed by ipg2016107

Javascript

<script>

// JavaScript program to implement
// the above approach

// Store the number of set bits at
// each position
let prefixCount = [[]];

// Function to precompute the prefix
// count array
function findPrefixCount(arr, size)
{

    // Iterate over the range[0, 31]
    for(let i = 0; i < 32; i++)
    {

        // Set the bit at position i
        // if arr[0] is set at position i
        prefixCount[i][0] = ((arr[0] >> i) & 1);

        // Traverse the array and
        // take prefix sum
        for(let j = 1; j < size; j++)
        {

            // Update prefixCount[i][j]
            prefixCount[i][j] = ((arr[j] >> i) & 1);
            prefixCount[i][j] += prefixCount[i][j - 1];
        }
    }
}

// Function to find the Bitwise AND
// of all array elements
function arrayBitwiseAND(size)
{

    // Stores the required result
    let result = 0;

    // Iterate over the range [0, 31]
    for(let i = 0; i < 32; i++)
    {

        // Stores the number of set
        // bits at position i
        let temp = prefixCount[i][size - 1];

        // If temp is N, then set ith
        // position in the result
        if (temp == size)
            result = (result | (1 << i));
    }

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

// Function to update the prefix count
// array in each query
function applyQuery(currentVal, newVal,
                       size)
{

    // Iterate through all the bits
    // of the current number
    for(let i = 0; i < 32; i++)
    {

        // Store the bit at position
        // i in the current value and
        // the new value
        let bit1 = ((currentVal >> i) & 1);
        let bit2 = ((newVal >> i) & 1);

        // If bit2 is set and bit1 is
        // unset, then increase the
        // set bits at position i by 1
        if (bit2 > 0 && bit1 == 0)
            prefixCount[i][size - 1]++;

        // If bit1 is set and bit2 is
        // unset, then decrease the
        // set bits at position i by 1
        else if (bit1 > 0 && bit2 == 0)
            prefixCount[i][size - 1]--;
    }
}

// Function to find the bitwise AND
// of the array after each query
function findbitwiseAND(queries, arr,
                           N, M)
{

    prefixCount = new Array(32);
    // Loop to create 2D array using 1D array
    for (var i = 0; i < prefixCount.length; i++) {
        prefixCount[i] = new Array(2);
    }

    // Fill the prefix count array
    findPrefixCount(arr, N);

    // Traverse the queries
    for(let i = 0; i < M; i++)
    {

        // Store the index and
        // the new value
        let id = queries[i][0];
        let newVal = queries[i][1];

        // Store the current element
        // at the index
        let currentVal = arr[id];

        // Update the array element
        arr[id] = newVal;

        // Apply the changes to the
        // prefix count array
        applyQuery(currentVal, newVal, N);

        // Print the bitwise AND of
        // the modified array
        arrayBitwiseAND(N);
    }
}

// Driver code

    let arr = [ 1, 2, 3, 4, 5 ];
    let queries = [[ 0, 2 ], [ 3, 3 ],
                        [ 4, 2 ]];
    let N = arr.length;
    let M = queries.length;

    findbitwiseAND(queries, arr, N, M);

// This code is contributed by code_hunt.
</script>

Output:

0 0 2

Time Complexity: O(N + M)
Auxiliary Space: O(N)