LeetCode #1598 — EASY

Crawler Log Folder

Build confidence with an intuition-first walkthrough focused on array fundamentals.

Solve on LeetCode

The Problem

Problem Statement

The Leetcode file system keeps a log each time some user performs a change folder operation.

The operations are described below:

"../" : Move to the parent folder of the current folder. (If you are already in the main folder, remain in the same folder).
"./" : Remain in the same folder.
"x/" : Move to the child folder named x (This folder is guaranteed to always exist).

You are given a list of strings logs where logs[i] is the operation performed by the user at the i^th step.

The file system starts in the main folder, then the operations in logs are performed.

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

Example 1:

Input: logs = ["d1/","d2/","../","d21/","./"]
Output: 2
Explanation: Use this change folder operation "../" 2 times and go back to the main folder.

Example 2:

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

Example 3:

Input: logs = ["d1/","../","../","../"]
Output: 0

Constraints:

1 <= logs.length <= 10³
2 <= logs[i].length <= 10
logs[i] contains lowercase English letters, digits, '.', and '/'.
logs[i] follows the format described in the statement.
Folder names consist of lowercase English letters and digits.

Step 01

Brute Force Baseline

Problem summary: The Leetcode file system keeps a log each time some user performs a change folder operation. The operations are described below: "../" : Move to the parent folder of the current folder. (If you are already in the main folder, remain in the same folder). "./" : Remain in the same folder. "x/" : Move to the child folder named x (This folder is guaranteed to always exist). You are given a list of strings logs where logs[i] is the operation performed by the user at the ith step. The file system starts in the main folder, then the operations in logs are performed. Return the minimum number of operations needed to go back to the main folder after the change folder operations.

Baseline thinking

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

Pattern signal: Array · Stack

Example 1

["d1/","d2/","../","d21/","./"]

Example 2

["d1/","d2/","./","d3/","../","d31/"]

Example 3

["d1/","../","../","../"]

Core Insight

What unlocks the optimal approach

Simulate the process but don’t move the pointer beyond the main folder.

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 #1598: Crawler Log Folder
class Solution {
    public int minOperations(String[] logs) {
        int ans = 0;
        for (var v : logs) {
            if ("../".equals(v)) {
                ans = Math.max(0, ans - 1);
            } else if (v.charAt(0) != '.') {
                ++ans;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1598: Crawler Log Folder
func minOperations(logs []string) int {
	ans := 0
	for _, v := range logs {
		if v == "../" {
			if ans > 0 {
				ans--
			}
		} else if v[0] != '.' {
			ans++
		}
	}
	return ans
}

# Accepted solution for LeetCode #1598: Crawler Log Folder
class Solution:
    def minOperations(self, logs: List[str]) -> int:
        ans = 0
        for v in logs:
            if v == "../":
                ans = max(0, ans - 1)
            elif v[0] != ".":
                ans += 1
        return ans

// Accepted solution for LeetCode #1598: Crawler Log Folder
impl Solution {
    pub fn min_operations(logs: Vec<String>) -> i32 {
        let mut depth = 0;
        for log in logs.iter() {
            if log == "../" {
                depth = (0).max(depth - 1);
            } else if log != "./" {
                depth += 1;
            }
        }
        depth
    }
}

// Accepted solution for LeetCode #1598: Crawler Log Folder
function minOperations(logs: string[]): number {
    let ans = 0;
    for (const x of logs) {
        if (x === '../') {
            ans && ans--;
        } else if (x !== './') {
            ans++;
        }
    }
    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.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

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.