feat: populate note files with problem descriptions and code stubs

Add populate-notes.mjs that fetches problem descriptions and
Python/C++ code stubs from LeetCode's GraphQL API. Populated
all 197 NeetCode 150 note files with:
- Problem description (examples, constraints)
- Python code stub (function signature)
- C++ code stub (function signature + includes)

API responses cached in leetcode/.cache/leetcode/ for instant re-runs.

org/study_deck_02/dsa/graphs/2924-find-champion-ii.org

@@ -1,18 +1,78 @@
 #+PROPERTY: STUDY_DECK_02
 * TODO 2924. Find Champion II :medium:
 :PROPERTIES:
-:NEETCODE: [[file:../../roadmap.org::*2924. Find Champion II][Roadmap]]
+:NEETCODE: [[file:../../roadmap.org::*2924. Find Champion II][2924. Find Champion II]]
 :END:
 
+There are ~n~ teams numbered from ~0~ to ~n - 1~ in a tournament; each team is also a node in a *DAG*.
+
+You are given the integer ~n~ and a *0-indexed* 2D integer array ~edges~ of length ~m~ representing the *DAG*, where ~edges[i] = [u_{i}, v_{i}]~ indicates that there is a directed edge from team ~u_{i}~ to team ~v_{i}~ in the graph.
+
+A directed edge from ~a~ to ~b~ in the graph means that team ~a~ is *stronger* than team ~b~ and team ~b~ is *weaker* than team ~a~.
+
+Team ~a~ will be the *champion* of the tournament if there is no team ~b~ that is *stronger* than team ~a~.
+
+Return /the team that will be the *champion* of the tournament if there is a *unique* champion, otherwise, return /~-1~/./
+
+*Notes*
+
+ - A *cycle* is a series of nodes ~a_{1}, a_{2}, ..., a_{n}, a_{n+1}~ such that node ~a_{1}~ is the same node as node ~a_{n+1}~, the nodes ~a_{1}, a_{2}, ..., a_{n}~ are distinct, and there is a directed edge from the node ~a_{i}~ to node ~a_{i+1}~ for every ~i~ in the range ~[1, n]~.
+
+ - A *DAG* is a directed graph that does not have any *cycle*.
+
+*Example 1:*
+
+
+#+begin_src
+Input: n = 3, edges = [[0,1],[1,2]]
+Output: 0
+Explanation: Team 1 is weaker than team 0. Team 2 is weaker than team 1. So the champion is team 0.
+#+end_src
+
+
+*Example 2:*
+
+
+#+begin_src
+Input: n = 4, edges = [[0,2],[1,3],[1,2]]
+Output: -1
+Explanation: Team 2 is weaker than team 0 and team 1. Team 3 is weaker than team 1. But team 1 and team 0 are not weaker than any other teams. So the answer is -1.
+#+end_src
+
+
+*Constraints:*
+
+ - ~1 <= n <= 100~
+
+ - ~m == edges.length~
+
+ - ~0 <= m <= n * (n - 1) / 2~
+
+ - ~edges[i].length == 2~
+
+ - ~0 <= edge[i][j] <= n - 1~
+
+ - ~edges[i][0] != edges[i][1]~
+
+ - The input is generated such that if team ~a~ is stronger than team ~b~, team ~b~ is not stronger than team ~a~.
+
+ - The input is generated such that if team ~a~ is stronger than team ~b~ and team ~b~ is stronger than team ~c~, then team ~a~ is stronger than team ~c~.
+
 ** TODO Approach
 Write your approach here.
 
 ** TODO Python
 #+begin_src python
-
+class Solution:
+    def findChampion(self, n: int, edges: List[List[int]]) -> int:
 #+end_src
 
 ** TODO C++
 #+begin_src cpp
-
+class Solution {
+public:
+    int findChampion(int n, vector<vector<int>>& edges) {
+        
+    }
+};
 #+end_src