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++

// 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;
}

Java

// 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.

Python3

# 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"

C#

// 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.

PHP

<?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
?>

Javascript

<script>

// Javascript 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)
{
    var 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
    var 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
var n = 80;
var l = 1, r = 4;
document.write( countUnsetBitsInGivenRange(n, l, r));

</script>

Output: 

4

Time Complexity: O(log n)
Space Complexity: O(1)