C, 33 bytes

I just saw the challenge today, but thought of a recursive solution.

B(n){return (n&1)+(n?B(n/2):0);};

C, 33 bytes

I just saw the challenge today, but thought of a recursive solution.

B(n){return (n&1)+(n?B(n/2):0);};

Edited to note that I assumed it was a smallest size challenge. The speed is linear to the number of bits required to contain the number. O(N) = k * Log2(N).

C, 33 bytes

I just saw the challenge today, but thought of a recursive solution.

B(n){return (n&1)+(n?B(n/2):0);};

C, 33 bytes

I just saw the challenge today, but thought of a recursive solution.

B(n){return (n&1)+(n?B(n/2):0);};