Count unset bits in a range

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

Examples:

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

Input : n = 80, 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. Let it be count.
  3. Calculate ans = (r – l + 1) – count.
  4. Return ans.

C++

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// C++ implementation to count unset 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 unset bits
// in the given range
unsigned int countUnsetBitsInGivenRange(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 unset bits in the range
    // 'l' to 'r' in 'n'
    return (r - l + 1) - countSetBits(n & num);
}
  
// Driver program to test above
int main()
{
    unsigned int n = 80;
    unsigned int l = 1, r = 4;
    cout << countUnsetBitsInGivenRange(n, l, r);
    return 0;
}

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Java

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// Java implementation to count unset 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 unset bits
    // in the given range
    static int countUnsetBitsInGivenRange(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 unset bits in the range
        // 'l' to 'r' in 'n'
        return (r - l + 1) - countSetBits(n & num);
    }
      
    // Driver code
    public static void main(String[] args)
    {
        int n = 80;
        int l = 1, r = 4;
          
        System.out.print(
            countUnsetBitsInGivenRange(n, l, r));
    }
}
  
// This code is contributed by Anant Agarwal.

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Python3

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# Python3 implementation to count 
# unset 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 += 1
    return count
  
# function to count unset bits
# in the given range
def countUnsetBitsInGivenRange (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 unset bits 
    # in the range 'l' to 'r' in 'n'
    return (r - l + 1) - countSetBits(n & num)
  
# Driver code to test above
n = 80
l = 1
r = 4
print(countUnsetBitsInGivenRange(n, l, r))
  
# This code is contributed by "Sharad_Bhardwaj"

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C#

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// C# implementation to count unset 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 unset bits
    // in the given range
    static int countUnsetBitsInGivenRange(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 unset bits in the range
        // 'l' to 'r' in 'n'
        return (r - l + 1) - countSetBits(n & num);
    }
      
    //Driver code
    public static void Main()
    {
        int n = 80;
        int l = 1, r = 4;
          
        Console.Write(countUnsetBitsInGivenRange(n, l, r));
    }
}
  
//This code is contributed by Anant Agarwal.

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PHP

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<?php
// php implementation to count 
// unset 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 unset 
// bits in the given range
function countUnsetBitsInGivenRange($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 unset
    // bits in the range
    // 'l' to 'r' in 'n'
    return ($r - $l + 1) - 
           countSetBits($n & $num);
}
  
    // Driver code
    $n = 80;
    $l = 1;
    $r = 4;
    echo countUnsetBitsInGivenRange($n, $l, $r);
  
// This code is contributed by mits 
?>

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Output:

4


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