Simplify Path Leetcode Problem 71 [Python Solution]

Question Link: Simplify Path LeetCode

Welcome back! Let’s Simplify Path.

In this blog post, we’re going to tackle the Simplify Path problem from LeetCode.

This problem falls under the category of Stack and is rated as a medium difficulty.

It’s a commonly asked question in technical interviews, and it’s a great exercise to strengthen your coding and problem-solving skills.

Problem Overview

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

In a Unix-style file system, the following conventions apply:
– A period ‘.’ refers to the current directory.
– A double period ‘..’ refers to the directory one level up.
– Multiple consecutive slashes (i.e., ‘//’) are treated as a single slash ‘/’.

For this problem, any other format of periods such as ‘…’ is treated as file or directory names.

The canonical path should adhere to 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, with no occurrences of the period ‘.’ or double period ‘..’.

Your goal is to return the simplified canonical path.

Example 1:
Input: path = “/home/”
Output: “/home”
Explanation: 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.

Understanding the Constraints

Before we dive into solving this problem, it’s essential to understand the constraints provided:

• 1 <= path.length <= 3000: The length of the input path is between 1 and 3000 characters.
• path consists of English letters, digits, period ‘.’, slash ‘/’, or underscore ‘_’.
• path is a valid absolute Unix path.

Now that we have a clear understanding of the problem, its requirements, and constraints, let’s proceed to solve it.

Simplify Path LeetCode Problem Solution

Let’s start by providing a Python solution to this problem.

def simplifyPath(path: str) -> str:
    stack = []
    for i in path.split('/'):
        if i == '..':
            if stack:
                stack.pop()
        elif i == '.' or i == '':
            continue
        else:
            stack.append(i)
    res = '/' + '/'.join(stack)
    return res

Reasoning Behind Our Approach

To simplify the given path, we utilize a stack data structure to keep track of the valid directories.

We parse the input path character by character, handling various cases:

1. If the character is not a slash ‘/’, we build the current file or directory’s name.
2. If we encounter a double dot “..” (indicating going up one directory level), we pop the top element from the stack if the stack is not empty.
3. If we encounter a single dot “.”, we simply skip it, as it represents the current directory.
4. We handle multiple consecutive slashes by skipping them.

Finally, we join the elements in our stack using slashes, ensuring a single slash at the beginning, and return the simplified canonical path.

Time and Space Complexity

Now, let’s discuss the time and space complexity of our solution:

Time Complexity: Our solution involves scanning the entire input path character by character, resulting in a time complexity of O(n), where n is the length of the path.

Space Complexity: We use a stack to keep track of directories, and in the worst case, this stack can store all the directories.

Hence, the space complexity is O(n).

Edge Cases a Valid Solution Must Consider

To handle this problem effectively, we must consider various edge cases.

These cases help ensure that our solution is robust and handles all possible scenarios:

1. Double Dots (“..”): When encountering “..” in the path, we need to navigate up one directory level if possible. This means we may need to pop elements from the stack.
2. Single Dots (“.”): When we encounter a single dot “.”, it indicates the current directory, so we should ignore it.
3. Multiple Consecutive Slashes: In a valid path, multiple consecutive slashes are treated as a single slash.
4. We need to ensure that we don’t add extra slashes to our simplified path.
5. Trailing Slash: The simplified path should not end with a trailing slash.

We need to handle this condition and ensure our result doesn’t include it.

By considering these edge cases, we can create a robust solution that works for a wide range of inputs.

Related Interview Questions By Company:

• Repeated Dna Sequences LeetCode
• Reverse Linked List II LeetCode
• Lowest Common Ancestor Of A Binary Search Tree LeetCode

Related Interview Questions By Difficulty:

• Simplify Path LeetCode
• Remove All Adjacent Duplicates In String II LeetCode
• Implement Stack Using Queues LeetCode

Related Interview Questions By Category:

• Kth Largest Element In A Stream LeetCode
• Trim A Binary Search Tree LeetCode
• Sort An Array LeetCode

Conclusion

In this blog post, we’ve explored the Simplify Path problem from LeetCode.

We provided a Python solution that uses a stack to efficiently handle the complexities of this problem.

By understanding the problem requirements, constraints, and edge cases, we’ve developed a solution that converts a given absolute path into its simplified canonical form.

We hope this post has been helpful to you, whether you’re preparing for technical interviews, learning data structures, or simply looking to solve interesting coding challenges.

If you found this content useful, please consider liking and commenting to support the our platform.

We also encourage you to ask questions, make suggestions, and share this content with others who might find it valuable.

Happy coding!