# Number of Steps to Reduce a Number in Binary Representation to One

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'

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
}
}```