Number of integers with odd number of set bits Number of integers with odd number of set bits Number of integers with odd number of set bits

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.

[sourcecode language=”CPP14″]
// 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;
}
[/sourcecode]

Output:

5


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