my solution for 1945

Author: jeremymanning

problems/1945/jeremymanning.md

@@ -1,11 +1,41 @@
 # [Problem 1945: Sum of Digits of String After Convert](https://leetcode.com/problems/sum-of-digits-of-string-after-convert/description/?envType=daily-question)
 
 ## Initial thoughts (stream-of-consciousness)
+- Another easy one 🥳!
+- Let's hard code in the lowercase letters (`letters = 'abcdefghijklmnopqrstuvwxyz'`)
+- Then we can make a `letter2num` decoder: `letter2num = {c: str(i + 1) for i, c in enumerate(letters)}`
+- Next, we just replace each letter with it's number, stitch it into a string, and convert to an integer: `num = int("".join([letter2num[c] for c in s]))`
+- Finally, loop through `k` times to sum the digits:
+```python
+for _ in range(k):
+    num = str(sum([int(i) for i in str(num)]))
+```
+- And then we just return `int(num)`
 
 ## Refining the problem, round 2 thoughts
+- There are no tricky edge cases, as far as I can tell; `k` is always at least 1, and the strings aren't super long so we don't need to worry about overflow issues
+- Let's try it!
+
 
 ## Attempted solution(s)
 ```python
-class Solution:  # paste your code here!
-    ...
+class Solution:
+    def getLucky(self, s: str, k: int) -> int:
+        letters = 'abcdefghijklmnopqrstuvwxyz'
+        letter2num = {c: str(i + 1) for i, c in enumerate(letters)}
+
+        num = int("".join([letter2num[c] for c in s]))
+        for _ in range(k):
+            num = str(sum([int(i) for i in str(num)]))
+
+        return int(num)
 ```
+- Given test cases pass
+- Let's try some other random ones:
+    - `s = "asdlkoifhawioefuh", k = 5`: pass
+    - `s = "dkfjkjdfijshdeweskjhaskljdfhjsldkjf", k = 4`: pass
+- I think we're good; submitting...
+
+![Screenshot 2024-09-02 at 8 41 31 PM](https://github.com/user-attachments/assets/0a1d3f63-c163-4bfb-a8ec-286ee3b9c0cc)
+
+Solved!