solutions and notes for problem 2582

Author: paxtonfitzpatrick

problems/2582/paxtonfitzpatrick.md

@@ -2,6 +2,38 @@
 
 ## Initial thoughts (stream-of-consciousness)
 
+- seems pretty straightforward
+- definitely going to be using modulo operator
+  - both for figuring out how many people from one end or the other who has the pillow and figuring out whether the current direction is forwards or backwards
+- actually, it's probably not optimal, but `itertools.cycle` could do this pretty easily
+  - let's try that approach first as a baseline
+  - trick here is to start with an iterable of `n` elements, but then extend it with items at indices `n-2` backwards through `1` so we don't repeat the first and last elements
+  - I think I can use `itertools.chain` to extend the iterable with a reversed partial copy like this without ever actually constructing it in memory.
+  - Also, `n` and/or `time` are really large, it might be better to use `itertools.islice` rather than acutally running the loop `time` times...
+
 ## Refining the problem
 
+- Let's try the non-`itertools` way now.
+  - first figure out how many people from one end or the other the pillow will be at time `time`
+  - then figure out whether we're currently going in the forward or backward direction
+
 ## Attempted solution(s)
+
+```python
+from itertools import chain, cycle, islice
+
+class Solution:
+    def passThePillow(self, n: int, time: int) -> int:
+        return next(islice(cycle(chain(range(1, n + 1), reversed(range(2, n)))), time, None))
+```
+
+![](https://github.com/paxtonfitzpatrick/leetcode-solutions/assets/26118297/07156d7a-fa66-49a7-b6fb-389acaf66907)
+
+```python
+class Solution:
+    def passThePillow(self, n: int, time: int) -> int:
+        offset = time % (n - 1)
+        return (n - offset) if time // (n - 1) % 2 else (offset + 1)
+```
+
+![](https://github.com/paxtonfitzpatrick/leetcode-solutions/assets/26118297/a96a6dc1-0852-4d3b-8c60-d1c997aea142)