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, 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

    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