My solution for 703

problems/703/jeremymanning.md

@@ -1,11 +1,115 @@
 # [Problem 703: Kth Largest Element in a Stream](https://leetcode.com/problems/kth-largest-element-in-a-stream/description/?envType=daily-question)
 
 ## Initial thoughts (stream-of-consciousness)
+- Kind of a fun one!
+- Inside the class, let's keep a running (sorted) list of the $k$ largest elements
+- When the class is initialized, we'll need to sort `nums` in descending order and set `self.k_largest` to the $k$ largest numbers (from largest to smallest)
+- When a number is added:
+    - If the number is less than or equal to the smallest number in `self.k_largest` (i.e., `self.k_largest[-1]`), return `self.k_largest[-1]`
+    - If the number is (strictly) greater than the smallest number in `self.k_largest`:
+        - Drop the current $k^\mathrm{th}$ largest number (`self.k_largest = self.k_largest[:-1]`)
+        - Loop through each element in the sequence until we find a place to insert `val`:
+        ```python
+        inserted = False
+        for i, x in enumerate(self.k_largest):
+            if x <= val:
+                self.k_largest = [*self.k_largest[:i], val, *self.k_largest[i:]]
+                inserted = True
+                break
+        if not inserted:
+            self.k_largest.append(val)
+        return self.k_largest[-1]
+        ```
 
 ## Refining the problem, round 2 thoughts
+- One thing I'm not sure of: are we guaranteed to get at least $k$ numbers when the class is intialized? I'll account for this (by appending -1's) just in case
+- Nothing particularly tricky here...let's just implement the solution
 
 ## Attempted solution(s)
 ```python
-class Solution:  # paste your code here!
-    ...
+class KthLargest:
+    def __init__(self, k: int, nums: List[int]):
+        self.k_largest = list(sorted(nums, reverse=True))[:k]
+        if len(self.k_largest) < k:
+            self.k_largest.extend([-1] * (k - len(nums)))
+
+    def add(self, val: int) -> int:
+        if val <= self.k_largest[-1]:
+            return self.k_largest[-1]
+        self.k_largest.pop()
+
+        inserted = False
+        for i, x in enumerate(self.k_largest):
+            if x <= val:
+                self.k_largest = [*self.k_largest[:i], val, *self.k_largest[i:]]
+                inserted = True
+                break
+        if not inserted:
+            self.k_largest.append(val)
+        return self.k_largest[-1]
 ```
+- The given test case passes
+- It's somewhat annoying to create new test cases, but let's see...
+    - `["KthLargest","add","add","add","add","add"], [[5,[4,5,8,2]],[3],[5],[10],[9],[4]]`: pass
+    - `["KthLargest","add","add","add","add","add","add","add"], [[6,[0,20,4,5,8,2]],[3],[21],[3],[5],[10],[9],[4]]`: pass
+- I think we're good...submitting...
+
+![Screenshot 2024-08-11 at 10 07 16 PM](https://github.com/user-attachments/assets/1c0e4751-5c7d-4c94-82e3-561630121aa4)
+
+🤦 Ah-- I misread the instructions (I thought we couldn't have negative numbers).  Easy fix...
+
+```python
+class KthLargest:
+    def __init__(self, k: int, nums: List[int]):
+        self.k_largest = list(sorted(nums, reverse=True))[:k]
+        if len(self.k_largest) < k:
+            self.k_largest.extend([-float("inf")] * (k - len(nums)))
+
+    def add(self, val: int) -> int:
+        if val <= self.k_largest[-1]:
+            return self.k_largest[-1]
+        self.k_largest.pop()
+
+        inserted = False
+        for i, x in enumerate(self.k_largest):
+            if x <= val:
+                self.k_largest = [*self.k_largest[:i], val, *self.k_largest[i:]]
+                inserted = True
+                break
+        if not inserted:
+            self.k_largest.append(val)
+        return self.k_largest[-1]
+```
+
+![Screenshot 2024-08-11 at 10 09 06 PM](https://github.com/user-attachments/assets/90a446aa-9d5e-4501-862a-c0658fa55603)
+
+Hah!  Did I win the prize for the slowest solution?  In any case, I'll take it: solved 🥳!
+
+- Note: a heap is definitely the way to solve this efficiently.  I could have used the built-in `heapq` (from the `heapq` module).
+- Let's see if I can get it to work quickly...
+
+```python
+import heapq
+
+class KthLargest:
+    def __init__(self, k: int, nums: List[int]):
+        self.k = k
+        self.k_largest = nums
+        heapq.heapify(self.k_largest)
+        while len(self.k_largest) > self.k:
+            heapq.heappop(self.k_largest)
+
+    def add(self, val: int) -> int:
+        heapq.heappush(self.k_largest, val)
+        while len(self.k_largest) > self.k:
+            heapq.heappop(self.k_largest)
+        return self.k_largest[0]
+```
+- That took a some hacking around based on the [heapq documentation](https://docs.python.org/3/library/heapq.html) but I got it to run...
+- Given test cases pass; submitting...
+
+![Screenshot 2024-08-11 at 10 18 27 PM](https://github.com/user-attachments/assets/737aaadb-8bfe-4c08-8a27-238d815fa96a)
+
+There we go-- quite a lot faster!
+
+