What is the time complexity of Crawler Log Folder?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Crawler Log Folder?

This problem uses Array, String, Stack. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Crawler Log Folder?

Insight 1: The problem can be solved using a stack-like approach or, more simply, by keeping track of the current depth in the directory structure. Insight 2: The order of operations matters. The number of `"../"` operations needed to return to the root depends on the net movement down the directory tree.

Which companies ask LeetCode 1598. Crawler Log Folder in interviews?

This problem is frequently asked at Atlassian, Flipkart, Mercari, ZS Associates.

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Crawler Log Folder - LeetCode 1598 Solution

Crawler Log Folder - Complete Solution Guide

Crawler Log Folder is LeetCode problem 1598, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 s

Detailed Explanation

The problem simulates navigating a file system. You're given a list of strings, `logs`, representing file system operations. Each string can be: * `"../"`: Move up one directory level. If already at the root, stay in place. * `"./"`: Stay in the current directory. * `"x/"`: Move down into the subdirectory named `x` (guaranteed to exist). The goal is to determine the minimum number of `"../"` operations needed to return to the root directory after performing all operations in `logs`.

Solution Approach

The solution uses a single integer variable, `depth`, to represent the current depth in the directory tree. It iterates through the `logs` array. For each log entry: * If it's `"../"`, it decrements `depth`, but `depth` cannot be less than 0 (we can't go above the root). * If it's `"./"`, it does nothing. * If it's a directory name, it increments `depth`. Finally, the value of `depth` represents the number of `"../"` operations needed to return to the root.

Step-by-Step Algorithm

Step 1: Initialize `depth` to 0 (representing the root directory).
Step 2: Iterate through each string in the `logs` array.
Step 3: If the string is `"../"`, decrement `depth`, ensuring it remains non-negative.
Step 4: If the string is `"./"`, skip to the next iteration.
Step 5: If the string is a directory name (neither `"../"` nor `"./"`), increment `depth`.
Step 6: After iterating through all logs, the final value of `depth` is the result.

Key Insights

Insight 1: The problem can be solved using a stack-like approach or, more simply, by keeping track of the current depth in the directory structure.
Insight 2: The order of operations matters. The number of `"../"` operations needed to return to the root depends on the net movement down the directory tree.
Insight 3: We can efficiently track the current depth using a single integer variable, avoiding the need for a more complex data structure such as a stack.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, String, Stack.

Companies

Asked at: Atlassian, Flipkart, Mercari, ZS Associates.

Crawler Log Folder | LeetCode 1598

Easy

Array String Stack

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int minOperations(String[] logs) {
        int depth = 0;
        for (String log : logs) {
            if (log.equals("../")) {
                depth = Math.max(0, depth - 1);
            } else if (!log.equals("./")) {
                depth++;
            }
        }
        return depth;
    }
}

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Problem Info

Description

Problem 1598: Crawler Log Folder - Detailed explanation and solution approaches.

Topics

Array String Stack

Companies

AtlassianFlipkartMercariZS Associates

Stats

Difficulty:Easy

Languages:4

Companies:4

Problem #:1598

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Step 1: Initialize `depth` to 0 (representing the root directory).
2
Step 2: Iterate through each string in the `logs` array.
3
Step 3: If the string is `"../"`, decrement `depth`, ensuring it remains non-negative.
4
Step 4: If the string is `"./"`, skip to the next iteration.
5
Step 5: If the string is a directory name (neither `"../"` nor `"./"`), increment `depth`.
6
Step 6: After iterating through all logs, the final value of `depth` is the result.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is linear because we process each log entry once. The space complexity is constant because we use only a single integer variable, regardless of the input size.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Using a Stack

A stack can be used to store the directory path. For each `"../"`, pop from the stack. For directory names, push onto the stack. The final stack size would be the result.

Time: O(n)Space: O(n) in the worst case (all directory names, no backtracks)

Trade-offs: The stack approach is less efficient in space complexity, potentially using more memory in the worst case, but conceptually clearer for those unfamiliar with direct depth tracking.

✨

Key Insights

1Insight 1: The problem can be solved using a stack-like approach or, more simply, by keeping track of the current depth in the directory structure.
2Insight 2: The order of operations matters. The number of `"../"` operations needed to return to the root depends on the net movement down the directory tree.
3Insight 3: We can efficiently track the current depth using a single integer variable, avoiding the need for a more complex data structure such as a stack.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Stack

LIFO structure - Last In, First Out. Like a stack of plates!

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Starting at the root and only moving down. The solution correctly returns the current depth.
Edge case 2: Starting at a subdirectory and moving up to the root then back down. The solution correctly handles the changing depth and gives the number of steps to reach the root after all operations.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the case where `depth` becomes negative when using `"../"`—the depth should be capped at 0 (the root directory).

✗Common mistake 2: Incorrectly handling `"./"`—this operation should not change the depth.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Simulating file system navigation in a command-line interface or graphical user interface.

🏦Real-world scenario 2: Tracking the number of steps needed to return to a base directory within a complex hierarchical system, like a version control system or an organization's internal network structure.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 6, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Crawler Log Folder"?

For this easy-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether string techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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