My solution to 2022

Author: jeremymanning

problems/2022/jeremymanning.md

@@ -1,11 +1,44 @@
 # [Problem 2022: Convert 1D Array Into 2D Array](https://leetcode.com/problems/convert-1d-array-into-2d-array/description/?envType=daily-question)
 
 ## Initial thoughts (stream-of-consciousness)
+- A nice and easy one
+- There are two clear approaches:
+    - We could build up the 2D array in a for loop
+    - Or we could use a list comprehension to build it in a single step
+- I'll go with the list comprehension approach, since I think we can do it as a one liner (plus a check that `m * n == len(original)`)
 
 ## Refining the problem, round 2 thoughts
+- Assuming the dimensions work out, I think we can just return `[original[x:(x + n)] for x in range(0, len(original), n)]`
 
 ## Attempted solution(s)
 ```python
-class Solution:  # paste your code here!
-    ...
+class Solution:
+    def construct2DArray(self, original: List[int], m: int, n: int) -> List[List[int]]:
+        if m * n == len(original):
+            return [original[x:(x + n)] for x in range(0, len(original), n)]
+        else:
+            return []
 ```
+- Given test cases pass
+- Submitting...
+
+![Screenshot 2024-08-31 at 9 48 44 PM](https://github.com/user-attachments/assets/f1d0b5ee-380f-4e7e-84f0-33e67ca02af5)
+
+- Solved!
+- Out of curiousity, what if I change the list comprehension to a generator?
+
+```python
+class Solution:
+    def construct2DArray(self, original: List[int], m: int, n: int) -> List[List[int]]:
+        if m * n == len(original):
+            return (original[x:(x + n)] for x in range(0, len(original), n))
+        else:
+            return []
+```
+
+![Screenshot 2024-08-31 at 9 50 35 PM](https://github.com/user-attachments/assets/e6e9819c-f3e8-4310-b491-953493db8550)
+
+- Interesting; runtime is slightly worse (though maybe still within the margin of error), but memory is slightly better.
+
+
+