[LeetCode] 71. Simplify Path

1. Question

Given a string path, which is an absolute path (starting with a slash '/') to a file or directory in a Unix-style file system, convert it to the simplified canonical path.

In a Unix-style file system, a period '.' refers to the current directory, a double period '..' refers to the directory up a level, and any multiple consecutive slashes (i.e. '//') are treated as a single slash '/'. For this problem, any other format of periods such as '...' are treated as file/directory names.

The canonical path should have the following format:

The path starts with a single slash '/'.
Any two directories are separated by a single slash '/'.
The path does not end with a trailing '/'.
The path only contains the directories on the path from the root directory to the target file or directory (i.e., no period '.' or double period '..')
Return the simplified canonical path.

Example 1:

Input: path = "/home/"
Output: "/home"
Explanation: Note that there is no trailing slash after the last directory name.

Example 2:

Input: path = "/../"
Output: "/"
Explanation: Going one level up from the root directory is a no-op, as the root level is the highest level you can go.

Example 3:

Input: path = "/home//foo/"
Output: "/home/foo"
Explanation: In the canonical path, multiple consecutive slashes are replaced by a single one.
 

2. Thoughts

1. Split the str with '/' (to remove duplicates)
2. use stack to pop when '..' is met
3. concat the result with / again. (Put / in the first for the root dir)

3. Tips learned

From :
https://ooeunz.tistory.com/7

3.1. Revisit Stack Pop

# Delete element from stack
 
stack = [1,2,3]
top = stack.pop()
 
print(top)
stack
 
# 3
# [1, 2]

4. My solution

class Solution:
    def simplifyPath(self, path: str) -> str:
        components = path.split("/")

        direc = []
        for component in components:
            if component == "." or component == "":
                continue
            elif component == "..":
                if len(direc) != 0:
                    direc.pop()
            else:
                direc.append(component)

        return "/" + "/".join(direc)

258/258 cases passed (28 ms)
Your runtime beats 97.44 % of python3 submissions
Your memory usage beats 57.54 % of python3 submissions (16.7 MB)

5. AI Solution and Improvements

Your solution for simplifying the Unix-style file path is quite effective. It correctly handles the key components of the problem: current directory (.), parent directory (..), and multiple consecutive slashes. Here's a breakdown of how your code works:

1. You split the path by slashes to get individual components.
2. You use a stack (direc) to keep track of directory names.
3. For each component, you handle the cases of '.', '..', and normal directory names.
4. Finally, you join the components in the stack with slashes and prepend a slash to form the simplified canonical path.

This is a clean and efficient approach. However, there's a minor enhancement you can make for clarity and performance:

• You can use list comprehension to filter out empty strings and '.' in one line instead of checking them inside the loop. This reduces the number of iterations and makes the code slightly more concise.

Here's the updated code with this enhancement:

class Solution:
    def simplifyPath(self, path: str) -> str:
        components = [component for component in path.split("/") if component not in ["", "."]]

        direc = []
        for component in components:
            if component == "..":
                if direc:
                    direc.pop()
            else:
                direc.append(component)

        return "/" + "/".join(direc)

This version of the code is a bit more streamlined but retains the same logic and efficiency as your original solution.