# 1404. Number of Steps to Reduce a Number in Binary Representation to One

## Description

Given the binary representation of an integer as a string s, 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 is guaranteed that you can always reach one for all test cases.

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'

## Solutions

Solution 1: Simulation

We simulate operations $1$ and $2$, while using carry to record the carry-over.

The time complexity is $O(n)$, where $n$ is the length of the string $s$. The space complexity is $O(1)$.

• class Solution {
public int numSteps(String s) {
boolean carry = false;
int ans = 0;
for (int i = s.length() - 1; i > 0; --i) {
char c = s.charAt(i);
if (carry) {
if (c == '0') {
c = '1';
carry = false;
} else {
c = '0';
}
}
if (c == '1') {
++ans;
carry = true;
}
++ans;
}
if (carry) {
++ans;
}
return ans;
}
}

• class Solution {
public:
int numSteps(string s) {
int ans = 0;
bool carry = false;
for (int i = s.size() - 1; i; --i) {
char c = s[i];
if (carry) {
if (c == '0') {
c = '1';
carry = false;
} else
c = '0';
}
if (c == '1') {
++ans;
carry = true;
}
++ans;
}
if (carry) ++ans;
return ans;
}
};

• class Solution:
def numSteps(self, s: str) -> int:
carry = False
ans = 0
for c in s[:0:-1]:
if carry:
if c == '0':
c = '1'
carry = False
else:
c = '0'
if c == '1':
ans += 1
carry = True
ans += 1
if carry:
ans += 1
return ans


• func numSteps(s string) int {
ans := 0
carry := false
for i := len(s) - 1; i > 0; i-- {
c := s[i]
if carry {
if c == '0' {
c = '1'
carry = false
} else {
c = '0'
}
}
if c == '1' {
ans++
carry = true
}
ans++
}
if carry {
ans++
}
return ans
}