Number of integers with odd number of set bits

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Given a number n, count number of integers smaller than or equal to n that have odd number of set bits.
Examples:

Input : 5
Output : 3
Explanation :
Integers with odd number of 
set bits in range 1 to 5 :
0 contains 0 set bits
1 contains 1 set bits
2 contains 1 set bits
3 contains 2 set bits
4 contains 1 set bits
5 contains 2 set bits

Input : 10
Output : 5
Explanation :
Integers with odd set bits are 1, 2,
4, 7 and 8.

Prerequisites: Count number of set bits
The idea is based on below fact.

If n is odd then there are total n+1 integers smaller than or equal to n (0, 1, 2 … n) and half of these integers contain odd number of set bits.

How to handle case when n is even? We know result for n-1. We count set bits in n and add 1 to n/2 if the count is odd. Else we return n/2.

C++

// CPP code to find numbers with
// odd number of set bits
#include <bits/stdc++.h>
using namespace std;
 
// function that returns the number
// of integers with odd number of
// set bits
int countWithOddSetBits(int n)
{
    // If n is odd, then half of the
    // integers in (0, 1, .. n) contain
    // odd number of set bits.
    if (n % 2 != 0)
        return (n + 1) / 2;
 
    // If n is even, we know result for
    // n-1. We explicitly compute set bit
    // count in n.
    int count = __builtin_popcount(n);
 
    int ans = n / 2;
    if (count % 2 != 0)
        ans++;
    return ans;
}
 
// Driver code
int main()
{
    int n = 10;
    cout << countWithOddSetBits(n);
    return 0;
}

C

// C code to find numbers with
// odd number of set bits
#include <stdio.h>
 
// function that returns the number
// of integers with odd number of
// set bits
int countWithOddSetBits(int n)
{
    // If n is odd, then half of the
    // integers in (0, 1, .. n) contain
    // odd number of set bits.
    if (n % 2 != 0)
        return (n + 1) / 2;
 
    // If n is even, we know result for
    // n-1. We explicitly compute set bit
    // count in n.
    int count = __builtin_popcount(n);
 
    int ans = n / 2;
    if (count % 2 != 0)
        ans++;
    return ans;
}
 
// Driver code
int main()
{
    int n = 10;
    printf("%d",countWithOddSetBits(n));
    return 0;
}
 
// This code is contributed by kothavvsaaash.

Java

// Java code to find numbers 
// with odd number of set bits
import java.io.*;
 
class GFG 
{
     
// function that returns the
// number of integers with 
// odd number of set bits
static int countWithOddSetBits(int n)
{
    // If n is odd, then half 
    // of the integers in 
    // (0, 1, .. n) contain
    // odd number of set bits.
    if (n % 2 != 0)
        return (n + 1) / 2;
 
    // If n is even, we know 
    // result for n-1. We 
    // explicitly compute set
    // bit count in n.
    int count = (n);
 
    int ans = n / 2;
    if (count % 2 != 0)
        ans++;
    return ans;
}
 
// Driver Code
public static void main (String[] args) 
{
    int n = 10;
    System.out.println( countWithOddSetBits(n));
}
}
 
// This code is contributed by aj_36

Python3

# Python 3 code to find numbers with
# odd number of set bits
 
# function that returns the number
# of integers with odd number of
# set bits
def countWithOddSetBits(n):
     
    # If n is odd, then half of the
    # integers in (0, 1, .. n) contain
    # odd number of set bits.
    if (n % 2 != 0):
        return (n + 1) / 2
 
    # If n is even, we know result for
    # n-1. We explicitly compute set 
    # bit count in n.
    count = bin(n).count('1')
 
    ans = n / 2
    if (count % 2 != 0):
        ans += 1
    return ans
 
# Driver code
if __name__ == '__main__':
    n = 10
    print(int(countWithOddSetBits(n)))
 
# This code is contributed by
# Surendra_Gangwar

C#

// C# code to find numbers 
// with odd number of set bits
using System;
 
class GFG
{
     
// function that returns the
// number of integers with 
// odd number of set bits
static int countWithOddSetBits(int n)
{
    // If n is odd, then half 
    // of the integers in 
    // (0, 1, .. n) contain
    // odd number of set bits.
    if (n % 2 != 0)
        return (n + 1) / 2;
 
    // If n is even, we know 
    // result for n-1. We 
    // explicitly compute set
    // bit count in n.
    int count = (n);
 
    int ans = n / 2;
    if (count % 2 != 0)
        ans++;
    return ans;
}
 
// Driver Code
static public void Main ()
{
    int n = 10;
    Console.WriteLine(countWithOddSetBits(n));
}
}
 
// This code is contributed by ajit

PHP

<?php
// PHP code to find numbers with
// odd number of set bits
 
// function that returns the number
// of integers with odd number of
// set bits
function countWithOddSetBits($n)
{
    // If n is odd, then half of 
    // the integers in (0, 1, .. n) 
    // contain odd number of set bits.
    if ($n % 2 != 0)
        return ($n + 1) / 2;
 
    // If n is even, we know result 
    // for n-1. We explicitly compute 
    // set bit count in n.
    $count = ($n);
 
    $ans = $n / 2;
    if ($count % 2 != 0)
        $ans++;
    return $ans;
}
 
// Driver code
$n = 10;
echo countWithOddSetBits($n);
 
// This code is contributed by aj_36
?>

Javascript

<script>
 
    // Javascript code to find numbers 
    // with odd number of set bits
     
    // function that returns the
    // number of integers with 
    // odd number of set bits
    function countWithOddSetBits(n)
    {
        // If n is odd, then half 
        // of the integers in 
        // (0, 1, .. n) contain
        // odd number of set bits.
        if (n % 2 != 0)
            return parseInt((n + 1) / 2, 10);
 
        // If n is even, we know 
        // result for n-1. We 
        // explicitly compute set
        // bit count in n.
        let count = (n);
 
        let ans = parseInt(n / 2, 10);
        if (count % 2 != 0)
            ans++;
        return ans;
    }
     
    let n = 10;
    document.write(countWithOddSetBits(n));
     
</script>

Output :

Time complexity: O(k), where k is the max number of bits in a number.
Auxiliary space: O(1)

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