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/advanced-graphs/0787-cheapest-flights-within-k-stops.org

@@ -1,18 +1,85 @@
 #+PROPERTY: STUDY_DECK_02
 * TODO 0787. Cheapest Flights Within K Stops :medium:
 :PROPERTIES:
-:NEETCODE: [[file:../../roadmap.org::*0787. Cheapest Flights Within K Stops][Roadmap]]
+:NEETCODE: [[file:../../roadmap.org::*0787. Cheapest Flights Within K Stops][0787. Cheapest Flights Within K Stops]]
 :END:
 
+There are ~n~ cities connected by some number of flights. You are given an array ~flights~ where ~flights[i] = [from_{i}, to_{i}, price_{i}]~ indicates that there is a flight from city ~from_{i}~ to city ~to_{i}~ with cost ~price_{i}~.
+
+You are also given three integers ~src~, ~dst~, and ~k~, return /*the cheapest price* from /~src~/ to /~dst~/ with at most /~k~/ stops. /If there is no such route, return/ /~-1~.
+
+*Example 1:*
+
+
+#+begin_src
+Input: n = 4, flights = [[0,1,100],[1,2,100],[2,0,100],[1,3,600],[2,3,200]], src = 0, dst = 3, k = 1
+Output: 700
+Explanation:
+The graph is shown above.
+The optimal path with at most 1 stop from city 0 to 3 is marked in red and has cost 100 + 600 = 700.
+Note that the path through cities [0,1,2,3] is cheaper but is invalid because it uses 2 stops.
+#+end_src
+
+
+*Example 2:*
+
+
+#+begin_src
+Input: n = 3, flights = [[0,1,100],[1,2,100],[0,2,500]], src = 0, dst = 2, k = 1
+Output: 200
+Explanation:
+The graph is shown above.
+The optimal path with at most 1 stop from city 0 to 2 is marked in red and has cost 100 + 100 = 200.
+#+end_src
+
+
+*Example 3:*
+
+
+#+begin_src
+Input: n = 3, flights = [[0,1,100],[1,2,100],[0,2,500]], src = 0, dst = 2, k = 0
+Output: 500
+Explanation:
+The graph is shown above.
+The optimal path with no stops from city 0 to 2 is marked in red and has cost 500.
+#+end_src
+
+
+*Constraints:*
+
+ - ~2 <= n <= 100~
+
+ - ~0 <= flights.length <= (n * (n - 1) / 2)~
+
+ - ~flights[i].length == 3~
+
+ - ~0 <= from_{i}, to_{i} < n~
+
+ - ~from_{i} != to_{i}~
+
+ - ~1 <= price_{i} <= 10^{4}~
+
+ - There will not be any multiple flights between two cities.
+
+ - ~0 <= src, dst, k < n~
+
+ - ~src != dst~
+
 ** TODO Approach
 Write your approach here.
 
 ** TODO Python
 #+begin_src python
-
+class Solution:
+    def findCheapestPrice(self, n: int, flights: List[List[int]], src: int, dst: int, k: int) -> int:
 #+end_src
 
 ** TODO C++
 #+begin_src cpp
-
+class Solution {
+public:
+    int findCheapestPrice(int n, vector<vector<int>>& flights, int src, int dst, int k) {
+        
+    }
+};
 #+end_src