Crawler Log Folder — Stack Depth Counter for Directory Navigation

Problem Statement#

The Leetcode file system starts in the main folder. You are given a list of strings logs where logs[i] is the operation performed by the user in the i-th step.

The operations are:

• "../" — move to the parent folder (stay at main if already there).
• "./" — stay in the current folder.
• "x/" — move to a child folder named x.

Return the minimum number of operations needed to go back to the main folder after all the operations.

Constraints:

• 1 <= logs.length <= 10^3
• 2 <= logs[i].length <= 10
• logs[i] contains lowercase English letters, digits, and/or '.' and '/'.
• It is guaranteed that logs[i] follows the described format.

Input:  logs = ["d1/","d2/","../","d21/","./"]
Output: 2
Explanation: Use "../" twice to return to main.

Input:  logs = ["d1/","d2/","./","d3/","../","d31/"]
Output: 3

Input:  logs = ["../","../","../"]
Output: 0
Explanation: Already at the main folder; "../" keeps you there.

Why This Problem Matters#

LC 1598 models a real problem every operating system solves: tracking the current directory depth in a hierarchical file system. The same logic underpins cd commands in Unix/Linux shells, file path canonicalization, and directory traversal in crawler bots (hence the name).

The problem looks like it requires a stack — and a full stack implementation works correctly — but the key insight is that you only need the depth, not the actual path. A single integer counter replaces the stack entirely, reducing space from O(n) to O(1).

Companies like Amazon and Google ask this problem to test whether candidates can abstract away unnecessary complexity: not every "stack-like" problem requires an explicit stack.

The Core Insight#

A file system path has one key property: the depth is always non-negative. When you navigate into a child folder, depth increases by 1. When you navigate to a parent, depth decreases by 1 (but never below 0). When you stay in place, depth does not change.

At the end of all operations, the answer is simply the current depth — that is the minimum number of "../" operations needed to reach the root.

Why a counter, not a full stack? You never need to know the actual folder names to answer this question. The only information needed is "how deep are we?" A full stack wastes space tracking names you never use.

When would you need a full stack? If the problem asked for the actual path after operations, or if you needed to detect cycles, you would need to store the folder names. Here, depth alone suffices.

Visual Dry Run#

Input: logs = ["d1/","d2/","../","d21/","./"]

Answer: depth = 2 → need 2 operations to return to root.

Input: logs = ["../","../","../"]

Answer: 0. Already at root, nothing to undo.

Solution (Optimal)#

# Python — O(n) time, O(1) space, depth counter only
def minOperations(logs: list[str]) -> int:
    depth = 0  # current depth from root (0 = at root)
 
    for log in logs:
        if log == '../':
            # Move to parent; cannot go above root
            depth = max(0, depth - 1)
        elif log == './':
            # Stay in current folder; no change
            pass
        else:
            # Move into a child folder
            depth += 1
 
    # The answer is the current depth —
    # that many "../" operations return to root
    return depth

// JavaScript — O(n) time, O(1) space
function minOperations(logs) {
    let depth = 0;
 
    for (const log of logs) {
        if (log === '../') {
            depth = Math.max(0, depth - 1);
        } else if (log === './') {
            // no-op: stay in current folder
        } else {
            depth++;  // enter child folder
        }
    }
 
    return depth;
}

Complexity:

Common Mistakes#

1. Going below 0 on "../". At the root, "../" keeps you at the root — depth cannot go negative. Use max(0, depth - 1) or an explicit if depth > 0: depth -= 1.

2. Not handling "./" as a no-op. Some candidates accidentally fall through to the else branch and increment depth for "./". Always check for "./" explicitly before the else.

3. Using a full stack when a counter suffices. Storing each folder name in a stack costs O(n) space unnecessarily. If the problem asks for the final path, you need the stack; if it only asks for the depth, a counter is optimal.

4. Comparing against the wrong string. The logs use "../" (with trailing slash), not "..". In Java, use .equals("../") not == "../" (reference equality for strings is unreliable).

5. Counting from 1 instead of 0. The root folder is depth 0, not depth 1. Starting from 1 will make every answer off by one.

Interview Tips#

• Demonstrate that you can recognize when an apparent stack problem does not need an explicit stack: "I could use a stack and push/pop folder names, but since I only need the depth — not the actual path — a counter is sufficient and uses O(1) space instead of O(n)."
• Trace through the boundary case ("../" at root) explicitly — interviewers always check this.
• Mention that the answer equals the final depth: "The minimum number of '../' operations to return to root is exactly the current depth."

Follow-up Questions#

1. Return the final path as a string. Use a full stack; push folder names on child operations, pop on "../", join with '/'.
2. Canonicalize a Unix path (LC 71). Same idea but handle absolute paths, leading slashes, empty segments, and "../" across the entire input path.
3. Minimum operations if you can also jump directly to any depth. You can jump to root in one operation — so the answer is min(1, depth) if depth > 0, else 0.
4. Handle symbolic links (folders that point elsewhere). Requires tracking full paths and resolving cycles — a much harder graph traversal problem.
5. Multi-user file system with concurrent operations. Each user has an independent depth tracker; shared state requires synchronization.

Key Takeaways#

• A counter replaces a stack when you only need the depth, not the actual path — this is the key optimization here.
• The root floor constraint (max(0, depth - 1)) prevents negative depth and correctly models "cannot go above root."
• "./" is a no-op — handle it explicitly before the else branch to avoid accidentally incrementing depth.
• The answer equals the final depth: it takes exactly that many "../" operations to return to the root.
• This problem is a microcosm of "stack vs counter" reasoning — recognizing when you need the full data structure versus just an aggregate value is a key interview skill.
• The same counter pattern applies to matching parentheses with one bracket type, net stack depth in call traces, and folder-depth analysis in crawlers.