• 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'

Solution: Simulation

Time complexity: O(n)
Space complexity: O(1)

// Author: Huahua
class Solution {
public:
  int numSteps(string s) {
    int ans = 0;
    int carry = 0;
    // The highest bit must be 1, 
    // process to the 2nd highest bit
    for (int i = s.length() - 1; i > 0; --i) {
      if (s[i] - '0' + carry == 1) {
        ans += 2; // odd: +1, even: / 2
        carry = 1; // always has a carry
      } else {
        ans += 1; // even: / 2
        // carray remains e.g. 1 + 1 = 10, or 0 + 0 = 00
      }
    }
    // If there is a carry, then it's even, one more step.
    return ans + carry;
  }
};

# Author: Huahua
class Solution:
  def numSteps(self, s: str) -> int:
    ans, carry = 0, 0
    for i in range(1, len(s)):
      if ord(s[-i]) - ord('0') + carry == 1:
        ans += 2
        carry = 1
      else:
        ans += 1
    return ans + carry

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