Crawler Log Folder - Stacks - Easy - LeetCode
💻 coding

Crawler Log Folder - Stacks - Easy - LeetCode

1 min read 241 words
1 min read
ShareWhatsAppPost on X
  • 1The Leetcode file system logs folder change operations, including moving to parent, remaining, or moving to a child folder.
  • 2To return to the main folder, the minimum number of operations is calculated based on the logs provided.
  • 3The solution uses a stack to track folder changes, with a time complexity of O(n) and space complexity of O(n).

AI-generated summary · May not capture all nuances

Key Insight
AskGif

"The Leetcode file system logs folder change operations, including moving to parent, remaining, or moving to a child folder."

Crawler Log Folder - Stacks - Easy - LeetCode

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.

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.

public class Solution {
 public int MinOperations(string[] logs) {
 var stack = new Stack<string>();
 for(int i=0;i<logs.Length;i++){
 switch(logs[i]){
 case "../":
 if(stack.Count()>0){
 stack.Pop(); 
 } 
 break;
 case "./":
 break;
 default:
 stack.Push(logs[i]);
 break;
 }
 }
 
 return stack.Count();
 }
}

Time Complexity: O(n)

Space Complexity: O(n)

Enjoyed this article?

Share it with someone who'd find it useful.

ShareWhatsAppPost on X

sumitc91

Published on 3 October 2020 · 1 min read · 241 words

Part of AskGif Blog · coding

You might also like