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Count set bits in a range
  • Difficulty Level : Expert
  • Last Updated : 09 Nov, 2020

Given a non-negative number n and two values l and r. The problem is to count the number of set bits in the range l to r in the binary representation of n, i.e, to count set bits from the rightmost lth bit to the rightmost rth bit.

Constraint: 1 <= l <= r <= number of bits in the binary representation of n.

Examples:

Input : n = 42, l = 2, r = 5
Output : 2
(42)10 = (101010)2
There are '2' set bits in the range 2 to 5.

Input : n = 79, l = 1, r = 4
Output : 4

Approach: Following are the steps:

  1. Calculate num = ((1 << r) – 1) ^ ((1 << (l-1)) – 1). This will produce a number num having r number of bits and bits in the range l to r are the only set bits.
  2. Count number of set bits in the number (n & num). Refer this post.

C++




// C++ implementation to count set bits in the
// given range
#include <bits/stdc++.h>
  
using namespace std;
  
// Function to get no of set bits in the
// binary representation of 'n'
unsigned int countSetBits(int n)
{
    unsigned int count = 0;
    while (n) {
        n &= (n - 1);
        count++;
    }
    return count;
}
  
// function to count set bits in the given range
unsigned int countSetBitsInGivenRange(unsigned int n,
                       unsigned int l, unsigned int r)
{
    // calculating a number 'num' having 'r' number
    // of bits and bits in the range l to r are the 
    // only set bits
    int num = ((1 << r) - 1) ^ ((1 << (l - 1)) - 1);
  
    // returns number of set bits in the range
    // 'l' to 'r' in 'n'
    return countSetBits(n & num);
}
  
// Driver program to test above
int main()
{
    unsigned int n = 42;
    unsigned int l = 2, r = 5;
    cout << countSetBitsInGivenRange(n, l, r);
    return 0;
}


Java




// Java implementation to count set bits in the
// given range
class GFG {
      
    // Function to get no of set bits in the
    // binary representation of 'n'
    static int countSetBits(int n)
    {
        int count = 0;
        while (n > 0) {
            n &= (n - 1);
            count++;
        }
          
        return count;
    }
  
    // function to count set bits in the given range
    static int countSetBitsInGivenRange(int n, int l, int r)
    {
          
        // calculating a number 'num' having 'r' number
        // of bits and bits in the range l to r are the
        // only set bits
        int num = ((1 << r) - 1) ^ ((1 << (l - 1)) - 1);
  
        // returns number of set bits in the range
        // 'l' to 'r' in 'n'
        return countSetBits(n & num);
    }
      
    // Driver code
    public static void main(String[] args)
    {
        int n = 42;
        int l = 2, r = 5;
          
        System.out.print(countSetBitsInGivenRange(n, l, r));
    }
}
  
// This code is contributed by Anant Agarwal.


Python3




# Python3 implementation to count
# set bits in the given range
  
# Function to get no of set bits in the
# binary representation of 'n'
def countSetBits(n):
    count = 0
    while (n):
        n &= (n - 1)
        count = count + 1
      
    return count
   
# function to count set bits in
# the given range
def countSetBitsInGivenRange(n, l,  r):
  
    # calculating a number 'num' having
    # 'r' number of bits and bits in the
    # range l to r are the  only set bits
    num = ((1 << r) - 1) ^ ((1 << (l - 1)) - 1)
   
    # returns number of set bits in the range
    # 'l' to 'r' in 'n'
    return countSetBits(n & num)
  
# Driver program to test above
n = 42
l = 2
r = 5
ans = countSetBitsInGivenRange(n, l, r)
print (ans)
  
# This code is contributed by Saloni Gupta.


C#




// C# implementation to count set bits in the
// given range
using System;
  
class GFG {
      
    // Function to get no of set bits in the
    // binary representation of 'n'
    static int countSetBits(int n)
    {
        int count = 0;
          
        while (n>0) {
            n &= (n - 1);
            count++;
        }
          
        return count;
    }
       
    // function to count set bits in the given range
    static int countSetBitsInGivenRange(int n,
                                       int l, int r)
    {
          
        // calculating a number 'num' having 'r' number
        // of bits and bits in the range l to r are the 
        // only set bits
        int num = ((1 << r) - 1) ^ ((1 << (l - 1)) - 1);
       
        // returns number of set bits in the range
        // 'l' to 'r' in 'n'
        return countSetBits(n & num);
    }
      
    //Driver code
    public static void Main()
    {
        int n = 42;
        int l = 2, r = 5;
          
        Console.WriteLine(countSetBitsInGivenRange(n, l, r));
    }
}
  
// This code is contributed by Anant Agarwal.


PHP




<?php
// PHP implementation to count
// set bits in the given range
  
// Function to get no of set bits in 
// the binary representation of 'n'
function countSetBits($n)
{
    $count = 0;
    while ($n)
    {
        $n &= ($n - 1);
        $count++;
    }
    return $count;
}
  
// function to count set 
// bits in the given range
function countSetBitsInGivenRange($n, $l, $r)
{
      
    // calculating a number 'num'
    // having 'r' number of bits
    // and bits in the range l to
    // r are the only set bits
    $num = ((1 << $r) - 1) ^ 
           ((1 << ($l - 1)) - 1);
  
    // returns number of 
    // set bits in the range
    // 'l' to 'r' in 'n'
    return countSetBits($n & $num);
}
  
    // Driver Code
    $n = 42;
    $l = 2;
    $r = 5;
    echo countSetBitsInGivenRange($n, $l, $r);
  
// This code is contributed by Ajit.
?>



Output:



2


This article is contributed by Ayush Jauhari. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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