Given a non-negative integer num, return the number of steps to reduce it to zero. If the current number is even, you have to divide it by 2, otherwise, you have to subtract 1 from it.
 
Example 1:
Input: num = 14
Output: 6
Explanation: 
Step 1) 14 is even; divide by 2 and obtain 7. 
Step 2) 7 is odd; subtract 1 and obtain 6.
Step 3) 6 is even; divide by 2 and obtain 3. 
Step 4) 3 is odd; subtract 1 and obtain 2. 
Step 5) 2 is even; divide by 2 and obtain 1. 
Step 6) 1 is odd; subtract 1 and obtain 0.

Example 2:
Input: num = 8
Output: 4
Explanation: 
Step 1) 8 is even; divide by 2 and obtain 4. 
Step 2) 4 is even; divide by 2 and obtain 2. 
Step 3) 2 is even; divide by 2 and obtain 1. 
Step 4) 1 is odd; subtract 1 and obtain 0.

Example 3:
Input: num = 123
Output: 12

 
Constraints:

0 <= num <= 10^6

Source: leetcode | Difficulty: EASY
Tags: Bit manipulation

Mathematics

Starter code: class Solution:
    def numberOfSteps (self, num: int) -> int:
        

Solutions:
class Solution:

	def numberOfSteps(self, num: int) -> int:
		steps = 0
		while num > 0:
			if num % 2 == 0:
				num /= 2
			else:
				num -= 1
			steps += 1
		return steps


class Solution:

	def numberOfSteps(self, num: int) -> int:
		bnum = bin(num)
		return len(bnum) + bnum.count('1') - 3


class Solution:

	def numberOfSteps(self, num: int) -> int:
		count = 0
		while num > 0:
			if num & 1:
				count += 1
			num = num >> 1
			count += 1
		return count - 1 if count > 0 else count