Introduction

In this blog post, we will explore how to solve the LeetCode problem "1598" 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

The Leetcode file system keeps a log each time some user performs a change folder operation. The operations are described below: "../" : Move to the parent folder of the current folder. (If you are already in the main folder, remain in the same folder). "./" : Remain in the same folder. "x/" : Move to the child folder named x (This folder is guaranteed to always exist). You are given a list of strings logs where logs[i] is the operation performed by the user at the ith step. The file system starts in the main folder, then the operations in logs are performed. Return the minimum number of operations needed to go back to the main folder after the change folder operations. Example 1: Input: logs = ["d1/","d2/","../","d21/","./"] Output: 2 Explanation: Use this change folder operation "../" 2 times and go back to the main folder. Example 2: Input: logs = ["d1/","d2/","./","d3/","../","d31/"] Output: 3 Example 3: Input: logs = ["d1/","../","../","../"] Output: 0 Constraints: 1 <= logs.length <= 103 2 <= logs[i].length <= 10 logs[i] contains lowercase English letters, digits, '.', and '/'. logs[i] follows the format described in the statement. Folder names consist of lowercase English letters and digits.

Explanation

• Simulate the directory changes: Maintain a counter representing the current directory depth.
  • Process each log entry: Increment the depth for child folders, decrement for parent folders (but not below zero), and ignore current directory notations.
  • Return the final depth: The final depth represents the number of ".." operations needed to return to the main folder.

• Time Complexity: O(n), where n is the number of logs. Space Complexity: O(1)

Code

    class Solution:
    def minOperations(self, logs: list[str]) -> int:
        depth = 0
        for log in logs:
            if log == "../":
                depth = max(0, depth - 1)
            elif log == "./":
                continue
            else:
                depth += 1
        return depth