# Solving Leetcode Interviews in Seconds with AI: Number of Steps to Reduce a Number in Binary Representation to One


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1404" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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" corresponds 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'  

	# Explanation
	*   **High-Level Approach:** Simulate the process directly on the binary string. If the last bit is '0', right-shift (divide by 2). If the last bit is '1', add 1 (which may involve carrying over).
*   Maintain a carry bit to handle additions. The number of steps is the number of iterations of this process. The algorithm continues until the string becomes "1".
*   **Complexity:** O(N), where N is the length of the input string. Space complexity is O(1) since we are modifying the string in place.

	
	# Code
	```python
	def numSteps(s: str) -> int:
    s = list(s)
    steps = 0
    carry = 0
    while len(s) > 1:
        if s[-1] == '0':
            s.pop()
            steps += 1
        else:
            steps += 1
            i = len(s) - 1
            while i >= 0:
                if s[i] == '0':
                    s[i] = '1'
                    carry = 0
                    break
                else:
                    s[i] = '0'
                    carry = 1
                i -= 1
            if carry == 1:
                s.insert(0, '1')
    return steps
	```