Given a number s in their binary representation. Return the number of steps to reduce it to 1 under the following rules:


If the current number is even, you have to divide it by 2.


If the current number is odd, you have to add 1 to it.


It's guaranteed that you can always reach to one for all testcases.
 
Example 1:
Input: s = "1101"
Output: 6
Explanation: "1101" corressponds to number 13 in their decimal representation.
Step 1) 13 is odd, add 1 and obtain 14. 
Step 2) 14 is even, divide by 2 and obtain 7.
Step 3) 7 is odd, add 1 and obtain 8.
Step 4) 8 is even, divide by 2 and obtain 4.  
Step 5) 4 is even, divide by 2 and obtain 2. 
Step 6) 2 is even, divide by 2 and obtain 1.  

Example 2:
Input: s = "10"
Output: 1
Explanation: "10" corressponds to number 2 in their decimal representation.
Step 1) 2 is even, divide by 2 and obtain 1.  

Example 3:
Input: s = "1"
Output: 0

 
Constraints:

1 <= s.length <= 500
s consists of characters '0' or '1'
s[0] == '1'

Source: leetcode | Difficulty: MEDIUM_HARD
Tags: String algorithms

Bit manipulation

Starter code: class Solution:
    def numSteps(self, s: str) -> int:
        

Solutions:
class Solution:

	def numSteps(self, s: str) -> int:
		(i, mid_zero) = (0, 0)
		for j in range(1, len(s)):
			if s[j] == '1':
				mid_zero += j - i - 1
				i = j
		if i == 0:
			return len(s) - 1
		return mid_zero + 1 + len(s)