Queries for bitwise AND in the index range [L, R] of the given array (original) (raw)

Last Updated : 26 Nov, 2021

Given an array arr[] of N and Q queries consisting of a range [L, R]. the task is to find the bit-wise AND of all the elements of in that index range.
Examples:

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

Naive approach: Iterate through the range and find bit-wise AND of all the numbers in that range. This will take O(n) time for each query.
Efficient approach: If we look at the integers as binary number, we can easily see that condition for ith bit of our answer to be set is that ith bit of all the integers in the range [L, R] should be set.
So, we will calculate prefix-count for each bit. We will use this to find the number of integers in the range with ith bit set. If it is equal to the size of the range then the ith bit of our answer will also be set.
Below is the implementation of the above approach:

C++ `

// C++ implementation of the approach #include <bits/stdc++.h> #define MAX 100000 #define bitscount 32 using namespace std;

// Array to store bit-wise // prefix count int prefix_count[bitscount][MAX];

// Function to find the prefix sum void findPrefixCount(int arr[], int n) {

// Loop for each bit
for (int i = 0; i < bitscount; i++) {

    // Loop to find prefix count
    prefix_count[i][0] = ((arr[0] >> i) & 1);
    for (int j = 1; j < n; j++) {
        prefix_count[i][j] = ((arr[j] >> i) & 1);
        prefix_count[i][j] += prefix_count[i][j - 1];
    }
}

}

// Function to answer query int rangeAnd(int l, int r) {

// To store the answer
int ans = 0;

// Loop for each bit
for (int i = 0; i < bitscount; i++) {
    // To store the number of variables
    // with ith bit set
    int x;
    if (l == 0)
        x = prefix_count[i][r];
    else
        x = prefix_count[i][r]
            - prefix_count[i][l - 1];

    // Condition for ith bit
    // of answer to be set
    if (x == r - l + 1)
        ans = (ans | (1 << i));
}

return ans;

}

// Driver code int main() { int arr[] = { 7, 5, 3, 5, 2, 3 }; int n = sizeof(arr) / sizeof(int);

findPrefixCount(arr, n);

int queries[][2] = { { 1, 3 }, { 4, 5 } };
int q = sizeof(queries) / sizeof(queries[0]);

for (int i = 0; i < q; i++)
    cout << rangeAnd(queries[i][0],
                     queries[i][1])
         << endl;

return 0;

}

Java

// Java implementation of the approach import java.io.*;

class GFG {

static int MAX = 100000; static int bitscount =32;

// Array to store bit-wise // prefix count static int [][]prefix_count = new int [bitscount][MAX];

// Function to find the prefix sum static void findPrefixCount(int arr[], int n) {

// Loop for each bit
for (int i = 0; i < bitscount; i++) 
{

    // Loop to find prefix count
    prefix_count[i][0] = ((arr[0] >> i) & 1);
    for (int j = 1; j < n; j++) 
    {
        prefix_count[i][j] = ((arr[j] >> i) & 1);
        prefix_count[i][j] += prefix_count[i][j - 1];
    }
}

}

// Function to answer query static int rangeAnd(int l, int r) {

// To store the answer
int ans = 0;

// Loop for each bit
for (int i = 0; i < bitscount; i++) 
{
    // To store the number of variables
    // with ith bit set
    int x;
    if (l == 0)
        x = prefix_count[i][r];
    else
        x = prefix_count[i][r]
            - prefix_count[i][l - 1];

    // Condition for ith bit
    // of answer to be set
    if (x == r - l + 1)
        ans = (ans | (1 << i));
}

return ans;

}

// Driver code public static void main (String[] args) {

int arr[] = { 7, 5, 3, 5, 2, 3 };
int n = arr.length;

findPrefixCount(arr, n);

int queries[][] = { { 1, 3 }, { 4, 5 } };
int q = queries.length;

for (int i = 0; i < q; i++)
            System.out.println (rangeAnd(queries[i][0],queries[i][1]));

} }

// This code is contributed by ajit.

Python3

Python3 implementation of the approach

import numpy as np

MAX = 100000 bitscount = 32

Array to store bit-wise

prefix count

prefix_count = np.zeros((bitscount,MAX));

Function to find the prefix sum

def findPrefixCount(arr, n) :

# Loop for each bit 
for i in range(0, bitscount) :
    
    # Loop to find prefix count 
    prefix_count[i][0] = ((arr[0] >> i) & 1); 
    
    for j in range(1, n) :
        prefix_count[i][j] = ((arr[j] >> i) & 1); 
        prefix_count[i][j] += prefix_count[i][j - 1];

Function to answer query

def rangeOr(l, r) :

# To store the answer 
ans = 0; 

# Loop for each bit 
for i in range(bitscount) :
    
    # To store the number of variables 
    # with ith bit set 
    x = 0; 
    
    if (l == 0) :
        x = prefix_count[i][r]; 
    else :
        x = prefix_count[i][r] - prefix_count[i][l - 1];
        
    # Condition for ith bit
    # of answer to be set
    if (x == r - l + 1) :
        ans = (ans | (1 << i)); 
        
return ans;

Driver code

if name == "main" :

arr = [ 7, 5, 3, 5, 2, 3 ]; 
n = len(arr);

findPrefixCount(arr, n); 

queries = [ [ 1, 3 ], [ 4, 5 ] ]; 

q = len(queries); 

for i in range(q) :
    print(rangeOr(queries[i][0], queries[i][1]));

This code is contributed by AnkitRai01

C#

// C# implementation of the approach using System;

class GFG {

static int MAX = 100000; static int bitscount =32;

// Array to store bit-wise // prefix count static int [,]prefix_count = new int [bitscount,MAX];

// Function to find the prefix sum static void findPrefixCount(int []arr, int n) {

// Loop for each bit 
for (int i = 0; i < bitscount; i++) 
{ 

    // Loop to find prefix count 
    prefix_count[i,0] = ((arr[0] >> i) & 1); 
    for (int j = 1; j < n; j++) 
    { 
        prefix_count[i,j] = ((arr[j] >> i) & 1); 
        prefix_count[i,j] += prefix_count[i,j - 1]; 
    } 
}

}

// Function to answer query static int rangeAnd(int l, int r) {

// To store the answer 
int ans = 0; 

// Loop for each bit 
for (int i = 0; i < bitscount; i++) 
{ 
    // To store the number of variables 
    // with ith bit set 
    int x; 
    if (l == 0) 
        x = prefix_count[i,r]; 
    else
        x = prefix_count[i,r] 
            - prefix_count[i,l - 1]; 

    // Condition for ith bit 
    // of answer to be set 
    if (x == r - l + 1) 
        ans = (ans | (1 << i)); 
} 

return ans;

}

// Driver code public static void Main (String[] args) {

int []arr = { 7, 5, 3, 5, 2, 3 }; 
int n = arr.Length; 

findPrefixCount(arr, n); 

int [,]queries = { { 1, 3 }, { 4, 5 } }; 
int q = queries.GetLength(0); 

for (int i = 0; i < q; i++) 
        Console.WriteLine(rangeAnd(queries[i,0],queries[i,1]));

} }

// This code contributed by Rajput-Ji

JavaScript

Time complexity for pre-computation is O(n) and each query can be answered in O(1)

Auxiliary Space: O(bitcount * MAX)