Queries for bitwise AND in the index range [L, R] of the given array
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 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 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# 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
<script>
// Javascript implementation of the approach
let MAX = 100000;
let bitscount =32;
// Array to store bit-wise
// prefix count
let prefix_count = new Array(bitscount);
for (let i = 0; i < bitscount; i++)
{
prefix_count[i] = new Array(MAX);
for (let j = 0; j < MAX; j++)
{
prefix_count[i][j] = 0;
}
}
// Function to find the prefix sum
function findPrefixCount(arr, n)
{
// Loop for each bit
for (let i = 0; i < bitscount; i++)
{
// Loop to find prefix count
prefix_count[i][0] = ((arr[0] >> i) & 1);
for (let 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
function rangeAnd(l, r)
{
// To store the answer
let ans = 0;
// Loop for each bit
for (let i = 0; i < bitscount; i++)
{
// To store the number of variables
// with ith bit set
let 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;
}
let arr = [ 7, 5, 3, 5, 2, 3 ];
let n = arr.length;
findPrefixCount(arr, n);
let queries = [ [ 1, 3 ], [ 4, 5 ] ];
let q = queries.length;
for (let i = 0; i < q; i++)
document.write(rangeAnd(queries[i][0],queries[i][1]) + "</br>");
</script>
Time complexity for pre-computation is O(n) and each query can be answered in O(1)
Auxiliary Space: O(bitcount * MAX)
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