Description
https://leetcode.com/problems/crawler-log-folder/
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 namedx
(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.
Explanation
Just having a list to base on the operation and add the folders visited. The length of list is the distance from the current directory to the main directory.
Python Solution
class Solution:
def minOperations(self, logs: List[str]) -> int:
visited = []
for log in logs:
if log == "../":
if not visited:
continue
visited.pop()
elif log == "./":
continue
else:
visited.append(log)
return len(visited)
- Time Complexity: O(N).
- Space Complexity: O(N).