Question

Can someone explains why Brian Kernighan’s algorithm takes O(log N) to count set bits (1s) in an integer. A simple implementation of this algorithm is below (in JAVA)

int count_set_bits(int n){
    int count = 0;
    while(n != 0){
        n &= (n-1);
        count++;
    }
    return count;
}

I understand how it works by clearing the rightmost set bit one by one until it becomes 0, but I just don’t know how we get O(log N).

Answer 1

This algorithm goes through as many iterations as there are set bits. So if we have a 32-bit word with only the high bit set, then it will only go once through the loop. In the worst case, it will pass once per bit. An integer n has log(n) bits, hence the worst case is O(log(n)). Here’s your code annotated at the important bits (pun intended):

  int count_set_bits(int n){
        int count = 0; // count accumulates the total bits set 
        while(n != 0){
            n &= (n-1); // clear the least significant bit set
            count++;
        }
  }