my solution to 1598

Author: jeremymanning

problems/1598/jeremymanning.md

@@ -1,11 +1,42 @@
 # [Problem 1598: Crawler Log Folder](https://leetcode.com/problems/crawler-log-folder/description/?envType=daily-question)
 
 ## Initial thoughts (stream-of-consciousness)
-
+  - We could solve this with a stack:
+      - Start with an empty stack (main folder)
+      - changing the directory pushes the new folder onto the top of the stack
+      - `'../'` pops the top folder of the stack (if empty, do nothing)
+      - `'./'` does nothing
+      - Then return the length of the stack at the end
+  - We can also see that it doesn't actually matter _which_ folders we visit.  So we can use less memory by just keeping track of the depth:
+      - If the next operation starts with `'..'` and `depth > 0` subtract one from the depth
+      - If the next operation is (`'./'`) do nothing
+      - Otherwise add one to the depth
+      
 ## Refining the problem, round 2 thoughts
+- I'm seeing that folders "consist of lowercase letters and digits"-- so they may not always start with `'d'` like in the examples
+- Also: testing for `depth > 0` should happen in a nested statement so that `'../'` with `depth == 0` doesn't get counted as a "directory"
+- Let's just go with this solution; it seems straightforward
 
 ## Attempted solution(s)
 ```python
-class Solution:  # paste your code here!
-    ...
+class Solution:
+    def minOperations(self, logs: List[str]) -> int:
+        depth = 0
+        for x in logs:
+            if x == '../':
+                if depth > 0:
+                    depth -= 1
+            elif x == './':
+                continue
+            else:
+                depth += 1
+        return depth
 ```
+- given test cases pass
+- verifying folders can start with other characters using `logs = ["e1/","d2/","../","d21/","./","../"]`: pass
+- submitting...
+
+- ![Screenshot 2024-07-09 at 9 39 46 PM](https://github.com/ContextLab/leetcode-solutions/assets/9030494/1d640a2b-6490-4c3f-a7f4-849e72885c6e)
+
+Solved!
+