Queries for bitwise OR in the index range [L, R] of the given array

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

Examples:

Input: arr[] = {1, 3, 1, 2, 3, 4}, q[] = {{0, 1}, {3, 5}}
Output:
3
7
1 OR 3 = 3
2 OR 3 OR 4 = 7

Input: arr[] = {1, 2, 3, 4, 5}, q[] = {{0, 4}, {1, 3}}
Output:
7
7

Naive approach: Iterate through the range and find bit-wise OR 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 any of 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 its greater than 0 then the ith bit of our answer will also be set.

Below is the implementation of the above approach:

C++

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// 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 rangeOr(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 != 0)
            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 << rangeOr(queries[i][0],
                        queries[i][1])
             << endl;
  
    return 0;
}

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Java

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// 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 rangeOr(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 != 0)
            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 (rangeOr(queries[i][0],queries[i][1]));
  
}
}
  
// This code is contributed by Tushil. 

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Python3

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# 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 != 0) :
            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

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C#

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// 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 rangeOr(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 != 0)
            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(rangeOr(queries[i,0],queries[i,1]));
  
}
}
  
// This code is contributed by 29AjayKumar

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PHP

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<?php 
  
// PHP implementation of the approach
$MAX= 100000;
$bitscount = 32;
  
// Array to store bit-wise
// prefix count
$prefix_count = array_fill(0,$bitscount,array_fill(0,$MAX,NULL));
  
// Function to find the prefix sum
function findPrefixCount(&$arr, $n)
{
    global $MAX,$bitscount,$prefix_count;
  
    // Loop for each bit
    for ($i = 0; $i < $bitscount; $i++)
    {
        // Loop to find prefix count
        $prefix_count[$i][0] = (($arr[0] >> $i) & 1);
        for ($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 rangeOr($l, $r)
{
    global $MAX,$bitscount,$prefix_count;
      
    // To store the answer
    $ans = 0;
  
    // Loop for each bit
    for ($i = 0; $i < $bitscount; $i++) 
    {
        // To store the number of variables
        // with ith bit set
          
        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 != 0)
            $ans = ($ans | (1 << $i));
    }
  
    return $ans;
}
  
    // Driver code
    $arr =array( 7, 5, 3, 5, 2, 3 );
    $n = sizeof($arr) / sizeof($arr[0]);
  
    findPrefixCount($arr, $n);
  
    $queries = array(array( 1, 3 ), array( 4, 5 ));
    $q = sizeof($queries) / sizeof($queries[0]);
  
    for ($i = 0; $i < $q; $i++)
        echo rangeOr($queries[$i][0],
                        $queries[$i][1])."\n";
  
    return 0;
      
    // This code is contributed by ChitraNayal
?>

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Output:

7
3

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



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