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 32using namespace std;// Array to store bit-wise// prefix countint prefix_count[bitscount][MAX];// Function to find the prefix sumvoid 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 queryint 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 codeint 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 countstatic int [][]prefix_count = new int [bitscount][MAX];// Function to find the prefix sumstatic 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 querystatic 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 codepublic 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 npMAX = 100000bitscount = 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 approachusing 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
<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> |
Output:
1 2
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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