LeetCode #388 — MEDIUM

Longest Absolute File Path

Move from brute-force thinking to an efficient approach using stack strategy.

The Problem

Problem Statement

Suppose we have a file system that stores both files and directories. An example of one system is represented in the following picture:

Here, we have dir as the only directory in the root. dir contains two subdirectories, subdir1 and subdir2. subdir1 contains a file file1.ext and subdirectory subsubdir1. subdir2 contains a subdirectory subsubdir2, which contains a file file2.ext.

In text form, it looks like this (with ⟶ representing the tab character):

dir
⟶ subdir1
⟶ ⟶ file1.ext
⟶ ⟶ subsubdir1
⟶ subdir2
⟶ ⟶ subsubdir2
⟶ ⟶ ⟶ file2.ext

If we were to write this representation in code, it will look like this: "dir\n\tsubdir1\n\t\tfile1.ext\n\t\tsubsubdir1\n\tsubdir2\n\t\tsubsubdir2\n\t\t\tfile2.ext". Note that the '\n' and '\t' are the new-line and tab characters.

Every file and directory has a unique absolute path in the file system, which is the order of directories that must be opened to reach the file/directory itself, all concatenated by '/'s. Using the above example, the absolute path to file2.ext is "dir/subdir2/subsubdir2/file2.ext". Each directory name consists of letters, digits, and/or spaces. Each file name is of the form name.extension, where name and extension consist of letters, digits, and/or spaces.

Given a string input representing the file system in the explained format, return the length of the longest absolute path to a file in the abstracted file system. If there is no file in the system, return 0.

Note that the testcases are generated such that the file system is valid and no file or directory name has length 0.

Example 1:

Input: input = "dir\n\tsubdir1\n\tsubdir2\n\t\tfile.ext"
Output: 20
Explanation: We have only one file, and the absolute path is "dir/subdir2/file.ext" of length 20.

Example 2:

Input: input = "dir\n\tsubdir1\n\t\tfile1.ext\n\t\tsubsubdir1\n\tsubdir2\n\t\tsubsubdir2\n\t\t\tfile2.ext"
Output: 32
Explanation: We have two files:
"dir/subdir1/file1.ext" of length 21
"dir/subdir2/subsubdir2/file2.ext" of length 32.
We return 32 since it is the longest absolute path to a file.

Example 3:

Input: input = "a"
Output: 0
Explanation: We do not have any files, just a single directory named "a".

Constraints:

1 <= input.length <= 10⁴
input may contain lowercase or uppercase English letters, a new line character '\n', a tab character '\t', a dot '.', a space ' ', and digits.
All file and directory names have positive length.

Step 01

Brute Force Baseline

Problem summary: Suppose we have a file system that stores both files and directories. An example of one system is represented in the following picture: Here, we have dir as the only directory in the root. dir contains two subdirectories, subdir1 and subdir2. subdir1 contains a file file1.ext and subdirectory subsubdir1. subdir2 contains a subdirectory subsubdir2, which contains a file file2.ext. In text form, it looks like this (with ⟶ representing the tab character): dir ⟶ subdir1 ⟶ ⟶ file1.ext ⟶ ⟶ subsubdir1 ⟶ subdir2 ⟶ ⟶ subsubdir2 ⟶ ⟶ ⟶ file2.ext If we were to write this representation in code, it will look like this: "dir\n\tsubdir1\n\t\tfile1.ext\n\t\tsubsubdir1\n\tsubdir2\n\t\tsubsubdir2\n\t\t\tfile2.ext". Note that the '\n' and '\t' are the new-line and tab characters. Every file and directory has a unique absolute path in the file system, which is the order of directories that must be opened

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Stack

Example 1

"dir\n\tsubdir1\n\tsubdir2\n\t\tfile.ext"

Example 2

"dir\n\tsubdir1\n\t\tfile1.ext\n\t\tsubsubdir1\n\tsubdir2\n\t\tsubsubdir2\n\t\t\tfile2.ext"

Example 3

"a"

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #388: Longest Absolute File Path
class Solution {
    public int lengthLongestPath(String input) {
        int i = 0;
        int n = input.length();
        int ans = 0;
        Deque<Integer> stack = new ArrayDeque<>();
        while (i < n) {
            int ident = 0;
            for (; input.charAt(i) == '\t'; i++) {
                ident++;
            }

            int cur = 0;
            boolean isFile = false;
            for (; i < n && input.charAt(i) != '\n'; i++) {
                cur++;
                if (input.charAt(i) == '.') {
                    isFile = true;
                }
            }
            i++;

            // popd
            while (!stack.isEmpty() && stack.size() > ident) {
                stack.pop();
            }

            if (stack.size() > 0) {
                cur += stack.peek() + 1;
            }

            // pushd
            if (!isFile) {
                stack.push(cur);
                continue;
            }

            ans = Math.max(ans, cur);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #388: Longest Absolute File Path
func lengthLongestPath(input string) int {
	i, n := 0, len(input)
	ans := 0
	var stk []int
	for i < n {
		ident := 0
		for ; input[i] == '\t'; i++ {
			ident++
		}

		cur, isFile := 0, false
		for ; i < n && input[i] != '\n'; i++ {
			cur++
			if input[i] == '.' {
				isFile = true
			}
		}
		i++

		// popd
		for len(stk) > 0 && len(stk) > ident {
			stk = stk[:len(stk)-1]
		}

		if len(stk) > 0 {
			cur += stk[len(stk)-1] + 1
		}

		// pushd
		if !isFile {
			stk = append(stk, cur)
			continue
		}

		ans = max(ans, cur)
	}
	return ans
}

# Accepted solution for LeetCode #388: Longest Absolute File Path
class Solution:
    def lengthLongestPath(self, input: str) -> int:
        i, n = 0, len(input)
        ans = 0
        stk = []
        while i < n:
            ident = 0
            while input[i] == '\t':
                ident += 1
                i += 1

            cur, isFile = 0, False
            while i < n and input[i] != '\n':
                cur += 1
                if input[i] == '.':
                    isFile = True
                i += 1
            i += 1

            # popd
            while len(stk) > 0 and len(stk) > ident:
                stk.pop()

            if len(stk) > 0:
                cur += stk[-1] + 1

            # pushd
            if not isFile:
                stk.append(cur)
                continue

            ans = max(ans, cur)

        return ans

// Accepted solution for LeetCode #388: Longest Absolute File Path
struct Solution;

struct Folder {
    level: usize,
    length: usize,
}

impl Solution {
    fn length_longest_path(input: String) -> i32 {
        let mut stack: Vec<Folder> = vec![];
        let mut folder_length = 0;
        let mut max = 0;
        for line in input.split('\n') {
            let s = line.chars();
            let mut level: usize = 0;
            let mut is_folder = true;
            for c in s {
                match c {
                    '\t' => {
                        level += 1;
                    }
                    '.' => {
                        is_folder = false;
                        break;
                    }
                    _ => {}
                }
            }
            let name = line[level..].to_string();
            while let Some(top) = stack.pop() {
                if top.level >= level {
                    folder_length -= top.length;
                } else {
                    stack.push(top);
                    break;
                }
            }
            if is_folder {
                let length = name.len() + 1;
                let folder = Folder { level, length };
                stack.push(folder);
                folder_length += length;
            } else {
                max = max.max(name.len() + folder_length);
            }
        }
        max as i32
    }
}

#[test]
fn test() {
    let input =
        "dir\n\tsubdir1\n\t\tfile1.ext\n\t\tsubsubdir1\n\tsubdir2\n\t\tsubsubdir2\n\t\t\tfile2.ext"
            .to_string();
    let res = 32;
    assert_eq!(Solution::length_longest_path(input), res);
}

// Accepted solution for LeetCode #388: Longest Absolute File Path
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #388: Longest Absolute File Path
// class Solution {
//     public int lengthLongestPath(String input) {
//         int i = 0;
//         int n = input.length();
//         int ans = 0;
//         Deque<Integer> stack = new ArrayDeque<>();
//         while (i < n) {
//             int ident = 0;
//             for (; input.charAt(i) == '\t'; i++) {
//                 ident++;
//             }
// 
//             int cur = 0;
//             boolean isFile = false;
//             for (; i < n && input.charAt(i) != '\n'; i++) {
//                 cur++;
//                 if (input.charAt(i) == '.') {
//                     isFile = true;
//                 }
//             }
//             i++;
// 
//             // popd
//             while (!stack.isEmpty() && stack.size() > ident) {
//                 stack.pop();
//             }
// 
//             if (stack.size() > 0) {
//                 cur += stack.peek() + 1;
//             }
// 
//             // pushd
//             if (!isFile) {
//                 stack.push(cur);
//                 continue;
//             }
// 
//             ans = Math.max(ans, cur);
//         }
//         return ans;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan left (or right) to find the next greater/smaller element. The inner scan can visit up to n elements per outer iteration, giving O(n²) total comparisons. No extra space needed beyond loop variables.

MONOTONIC STACK

O(n) time

O(n) space

Each element is pushed onto the stack at most once and popped at most once, giving 2n total operations = O(n). The stack itself holds at most n elements in the worst case. The key insight: amortized O(1) per element despite the inner while-loop.

Shortcut: Each element pushed once + popped once → O(n) amortized. The inner while-loop does not make it O(n²).

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Breaking monotonic invariant

Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.

Usually fails on: Indices point to blocked elements and outputs shift.

Fix: Pop while invariant is violated before pushing current element.