Submitting my solution for 2582

Author: jeremymanning

problems/2582/jeremymanning.md

@@ -1,7 +1,35 @@
 # [Problem 2582: Pass the Pillow](https://leetcode.com/problems/pass-the-pillow)
 
 ## Initial thoughts (stream-of-consciousness)
+  - I think this is pretty straightforward
+  - For a list of length $n$, it takes $n - 1$ passes to get to the last person, and another $n - 1$ passes (backwards through the line) to get back to the first person.  So the cycle resets every $2(n - 1)$ passes.
+  - Therefore we can immeidately mod the number of passes by $2(n - 1)$ without affecting the final position
+  - After doing so, if the remainder is less than $n$, return the remainder + 1.  Otherwise (the remainder must be greater than or equal to $n$ and less than $2(n - 1) - 1$ return $2n$ - the remainder - 1.
 
 ## Refining the problem
+  - Any special cases to deal with?
+    - If $n == 1$, should we just return 1?  --> It looks like $2 \leq n \leq 1000$, so we don't need to handle this case
+    - If time is 0, just return 1?  --> Again, it looks like $1 \leq \mathrm{time} \leq 1000$, so no need to handle this case either
 
 ## Attempted solution(s)
+
+```python
+class Solution:
+    def passThePillow(self, n: int, time: int) -> int:
+        time %= 2 * (n - 1)
+        if time < n:
+            return time + 1
+        else:
+            return 2 * n - time - 1
+```
+- given test cases pass
+- new test cases:
+  - n = 1000, time = 1000: pass
+  - n = 50, time = 324: pass
+  - n = 4, time = 999: pass
+  - n = 2, time = 1000: pass
+- There don't seem to be any important edge cases that I'm missing; submitting...
+
+<img width="690" alt="Screenshot 2024-07-05 at 11 11 29 PM" src="https://github.com/ContextLab/leetcode-solutions/assets/9030494/163e0790-5c4a-408d-9fca-3ce769b83c4e">
+
+Solved!