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.

Input:s = "1101"Output:6Explanation:"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.

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

Input:s = "1"Output:0

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

Solution:

class Solution { // 1101 // 1110 // 111 // 1000 // 100 // 10 // 1 public int numSteps(String s) { int res = 0; int carry = 0; for (int i = s.length() - 1; i > 0; i--){ if (s.charAt(i) - '0' + carry == 0){ res++; carry = 0; }else if (s.charAt(i) - '0' + carry == 1) { res += 2; // after adding 1 and / 2 carry = 1; // example: 1 + 1 -> 10 -> 10 >> 1 -> carry 1 over }else { // both digit and carry is 1 res++; // current digit is 1, adding carry over 1 is 10 -> only do >> 1 and carry 1 over to the left carry = 1; } } return res + carry; // need to consider 1 at index 0 } }