LeetCode 388: Longest Absolute File Path Solution in Python – A Step-by-Step Guide

What Is LeetCode 388: Longest Absolute File Path?

LeetCode 388: Longest Absolute File Path provides a string representing a file system, where directories and files are listed with newline (\n) separators and tabs (\t) indicating nesting levels. Your task is to return the length of the longest absolute path to a file (including separators). It’s a problem that tests your ability to parse structured text and compute hierarchical lengths!

Problem Statement

• Input:
• input: String representing the file system (e.g., "dir\n\tsubdir1\n\t\tfile.ext").
• Output: Integer - Length of the longest absolute path to a file, or 0 if none.
• Rules:
• \n separates entries (directories or files).
• \t indicates nesting level (one tab = one level deeper).
• File has an extension (contains '.'), directory does not.
• Path length includes separators ('/') between levels.

Constraints

• 1 <= input.length <= 10⁴
• Input contains letters, digits, '\t', '\n', '.', '-'.
• No leading/trailing '\t' or '\n'.

Examples

• Input: input = "dir\n\tsubdir1\n\tsubdir2\n\t\tfile.ext"
• Output: 20
• Paths: "dir/subdir1", "dir/subdir2", "dir/subdir2/file.ext".
• Longest: "dir/subdir2/file.ext" = 3 + 1 + 7 + 1 + 8 = 20.
• Input: input = "dir\n\tsubdir1\n\t\tfile1.ext\n\t\tsubsubdir1\n\tsubdir2\n\t\tsubsubdir2\n\t\t\tfile2.ext"
• Output: 32
• Longest: "dir/subdir2/subsubdir2/file2.ext" = 32.
• Input: input = "a"
• Output: 0
• No file (no '.').

These show the path length puzzle—let’s solve it!

Understanding the Problem: Finding the Longest File Path

To tackle LeetCode 388 in Python, we need to: 1. Parse the string into directories and files based on \n and \t. 2. Compute the absolute path length to each file. 3. Track the maximum length efficiently.

A naive approach might build all paths and check, but that’s inefficient. We’ll use:

• Best Solution: Stack—O(n) time, O(n) space—fast and iterative.
• Alternative Solution: Dictionary—O(n) time, O(n) space—intuitive but slightly more complex.

Let’s navigate with the best solution!

Best Solution: Stack

Why This Is the Best Solution

The stack-based approach runs in O(n) time (n is string length) with O(n) space by processing the string in one pass, using a stack to track directory depths and lengths. It’s the most efficient—computing path lengths incrementally and avoiding redundant storage, making it both fast and clear!

How It Works

Think of it like a file explorer:

• Stack: Stores (length, level) pairs for each directory/file.
• Process:
• Split string by \n.
• Count tabs for level, compute name length.
• Pop stack until level matches, add new length.
• If file (contains '.'), update max length.
• Result: Max length or 0 if no files.

It’s like climbing directories and measuring paths!

Step-by-Step Example

For input = "dir\n\tsubdir1\n\tsubdir2\n\t\tfile.ext":

• Init: Stack = [], max_len = 0.
• "dir":
• Level = 0, Len = 3, Stack = [(3, 0)].
• "\tsubdir1":
• Level = 1, Len = 7, Stack = [(3, 0), (3+1+7=11, 1)].
• "\tsubdir2":
• Level = 1, Pop to level 0, Stack = [(3, 0)].
• Add: Stack = [(3, 0), (3+1+7=11, 1)].
• "\t\tfile.ext":
• Level = 2, Len = 8, Stack = [(3, 0), (11, 1), (11+1+8=20, 2)].
• File ('.'), max_len = 20.
• Result: 20.

For input = "a":

• "a": Stack = [(1, 0)], no '.', max_len = 0.
• Result: 0.

Code with Explanation

Here’s the Python code using a stack:

Copy codeclass Solution:
    def lengthLongestPath(self, input: str) -> int:
        # Stack to store (length, level) pairs
        stack = []
        max_len = 0

        # Split by newline
        for path in input.split('\n'):
            # Count tabs (level) and get name length
            level = 0
            while path[level] == '\t':
                level += 1
            name = path[level:]
            curr_len = len(name)

            # Pop stack until at parent level
            while stack and stack[-1][1] >= level:
                stack.pop()

            # Compute total length with parent
            total_len = stack[-1][0] + 1 + curr_len if stack else curr_len

            # If file, update max_len
            if '.' in name:
                max_len = max(max_len, total_len)
            else:
                stack.append((total_len, level))

        return max_len

Explanation

• Line 4: stack: List of (length, level) tuples.
• Line 5: max_len: Track longest file path.
• Line 7: Split by \n—O(n).
• Lines 9-13: For each path:
• Count tabs for level, get name—O(k) per path.
• Lines 15-16: Pop stack to match level—O(1) amortized.
• Lines 18-23:
• Compute total length with parent (+1 for '/')—O(1).
• If file ('.'), update max_len; else push to stack—O(1).
• Line 25: Return max_len—O(1).
• Time: O(n)—single pass, amortized O(1) per operation.
• Space: O(n)—stack size up to max depth.

This is like a file path builder—swift and smart!

Alternative Solution: Dictionary

Why an Alternative Approach?

The dictionary solution also runs in O(n) time with O(n) space by mapping levels to cumulative lengths, processing the string in one pass. It’s less intuitive—great for understanding a level-based approach, though it uses more space and is slightly more complex than the stack!

How It Works

• Dict: Map level to current path length.
• Process:
• Split by \n, count tabs.
• Update level length based on parent.
• Track max file length.
• Result: Max length or 0.

Step-by-Step Example

For input = "dir\n\tsubdir1\n\tsubdir2\n\t\tfile.ext":

• Init: Dict = {0: 0}, max_len = 0.
• "dir":
• Level = 0, Len = 3, Dict = {0: 3}.
• "\tsubdir1":
• Level = 1, Len = 7, Dict = {0: 3, 1: 3+1+7=11}.
• "\tsubdir2":
• Level = 1, Len = 7, Dict = {0: 3, 1: 11}.
• "\t\tfile.ext":
• Level = 2, Len = 8, Total = 11+1+8=20, max_len = 20.
• Result: 20.

Code with Explanation

Copy codeclass Solution:
    def lengthLongestPath(self, input: str) -> int:
        # Dict to store level -> length
        level_len = {0: 0}
        max_len = 0

        # Split by newline
        for path in input.split('\n'):
            # Count tabs and get name
            level = path.count('\t')
            name = path[level:]
            curr_len = len(name)

            # Compute total length with parent
            total_len = level_len[level - 1] + 1 + curr_len if level > 0 else curr_len

            # If file, update max_len
            if '.' in name:
                max_len = max(max_len, total_len)
            else:
                level_len[level] = total_len

        return max_len

Explanation

• Line 4: level_len: Dict mapping level to length, {0: 0} for root.
• Line 5: max_len: Track longest file path.
• Line 7: Split by \n—O(n).
• Lines 9-13:
• Count tabs, get name—O(k).
• Lines 15-19:
• Total length = parent + 1 + current—O(1).
• If file, update max_len; else store level length—O(1).
• Line 21: Return max_len—O(1).
• Time: O(n)—single pass.
• Space: O(n)—dict size up to max depth.

It’s a level tracker—works but bulkier!

Comparing the Two Solutions

• Stack (Best):
• Time: O(n)—linear.
• Space: O(n)—stack.
• Pros: Fast, clear, minimal overhead.
• Cons: Stack management.
• Dictionary (Alternative):
• Time: O(n)—linear.
• Space: O(n)—dict.
• Pros: Level-based clarity.
• Cons: More space, less intuitive.

Stack wins for simplicity and efficiency!

Additional Examples and Edge Cases

• "file.ext": Both return 8.
• "a\n\tb": Both return 0 (no file).
• Deep nesting: Stack scales cleanly.

Both handle these, stack is leaner.

Complexity Breakdown

• Stack:
• Time: O(n).
• Space: O(n).
• Dictionary:
• Time: O(n).
• Space: O(n).

Stack excels for clarity and execution.

Key Takeaways

• Stack: Build paths iteratively—fast and smart!
• Dictionary: Track levels—clear but heavier!
• Paths: Nesting drives solutions.
• Python Tip: Lists rock—see [Python Basics](/python/basics).

Final Thoughts: Find That Path

LeetCode 388: Longest Absolute File Path in Python is a parsing challenge. Stack-based is the efficiency champ, while dictionary offers a level-based alternative. Want more? Try LeetCode 394: Decode String or LeetCode 257: Binary Tree Paths. Ready to explore? Visit Solve LeetCode 388 on LeetCode and measure that path today!