From 580e6c12f1dc2f8998b75594a6c2d3d52be93665 Mon Sep 17 00:00:00 2001
From: "Jeremy R. Manning" <jeremymanning@users.noreply.github.com>
Date: Sat, 17 Aug 2024 23:40:58 -0400
Subject: [PATCH] My solution to 264

---
 problems/264/jeremymanning.md | 48 +++++++++++++++++++++++++++++++++--
 1 file changed, 46 insertions(+), 2 deletions(-)

diff --git a/problems/264/jeremymanning.md b/problems/264/jeremymanning.md
index eafbc09..1696e6f 100644
--- a/problems/264/jeremymanning.md
+++ b/problems/264/jeremymanning.md
@@ -1,11 +1,55 @@
 # [Problem 264: Ugly Number II](https://leetcode.com/problems/ugly-number-ii/description/?envType=daily-question)
 
 ## Initial thoughts (stream-of-consciousness)
+- One way to solve this would be:
+    - Start the sequence with 1
+    - Each subsequent element is generated by multiplying a previous element by 2, 3, or 5
+    - We could have 3 indices/pointers to keep track of the next thing to multiply by 2/3/5 (initially these should all point to the first element, which is 1).  Let's call these `i2`, `i3`, and `i5`.
+    - Now we just:
+        - Take the current 2/3/5 pointers' numbers and multiply by 2/3/5
+        - Take the minimum of the results (this occurs for pointer `i`).  Note: it's possible that *multiple* pointers could match-- e.g., if `i2 == 3` and `i3 == 2` then both the `i2` and `i3` results will be 6.
+        - Add it to the list and increment the matching pointer(s) (by 1).
+        - Repeat until we've gotten `n` numbers in the sequence and then return the last number in the sequence
 
 ## Refining the problem, round 2 thoughts
+- Let's see if this works...
 
 ## Attempted solution(s)
 ```python
-class Solution:  # paste your code here!
-    ...
+class Solution:
+    def nthUglyNumber(self, n: int) -> int:
+        ugly_numbers = [0] * n
+        ugly_numbers[0] = 1
+
+        i2, i3, i5 = 0, 0, 0
+        next2, next3, next5 = 2, 3, 5
+
+        for i in range(1, n):
+            x = min(next2, next3, next5)
+            ugly_numbers[i] = x
+
+            if x == next2:
+                i2 += 1
+                next2 = ugly_numbers[i2] * 2
+
+            if x == next3:
+                i3 += 1
+                next3 = ugly_numbers[i3] * 3
+
+            if x == next5:
+                i5 += 1
+                next5 = ugly_numbers[i5] * 5
+
+        return ugly_numbers[n - 1]
 ```
+- Given text cases pass
+- Let's try some others:
+    - `n = 1690`: pass
+    - `n = 250`: pass
+- Ok, it's probably good; submitting...
+
+![Screenshot 2024-08-17 at 11 39 32 PM](https://github.com/user-attachments/assets/a221898c-7840-4834-a190-41047867352b)
+
+Woah, apparently this was the way to solve it!
+
+