How do you solve Crawler Log Folder on LeetCode?

Crawler Log Folder (LeetCode #1598) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Each '../' operation decreases depth, while each 'x/' increases it.

What is the brute force solution for Crawler Log Folder?

The brute force approach for Crawler Log Folder is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach simulates each operation step-by-step. We can keep track of how many folders we move into and how many we move back, ultimately calculating how many operations are needed to return to the main folder. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Crawler Log Folder?

Crawler Log Folder uses the following concepts: Array, String, Stack. The recommended patterns to study are: Stack, Array.

What are the interview tips for Crawler Log Folder?

Clarify the problem requirements and constraints before diving into the solution. Explain your thought process as you code to demonstrate your understanding. Test your solution with edge cases, such as only '../' operations.

What are common mistakes when solving Crawler Log Folder?

Not using max(0, depth - 1) to prevent negative depth. Confusing the operations and miscounting the depth.

#1598

Crawler Log Folder

Easy

ArrayStringStackStackArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

The optimal solution also tracks the depth of the current folder but does so in a single pass through the logs. This approach is efficient and straightforward, ensuring we only count the necessary operations to return to the main folder.

⚙️

Algorithm

4 steps

1Step 1: Initialize a variable to track the current depth, starting at 0.
2Step 2: Loop through each log entry in the logs.
3Step 3: For each log, adjust the depth: decrease for '../', increase for 'x/', and ignore './'.
4Step 4: After processing all logs, the depth variable will directly represent the number of operations needed to return to the main folder.

solution.py8 lines

1def minOperations(logs):
2    depth = 0
3    for log in logs:
4        if log == '../':
5            depth = max(0, depth - 1)
6        elif log != './':
7            depth += 1
8    return depth

ℹ

Complexity note: The time complexity is O(n) since we only make a single pass through the logs. The space complexity is O(1) because we only use a single integer to track the depth.

1Each '../' operation decreases depth, while each 'x/' increases it.
2The './' operation does not affect the depth.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.