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/greedy/0134-gas-station.org

@@ -1,18 +1,73 @@
 #+PROPERTY: STUDY_DECK_02
 * TODO 0134. Gas Station :medium:
 :PROPERTIES:
-:NEETCODE: [[file:../../roadmap.org::*0134. Gas Station][Roadmap]]
+:NEETCODE: [[file:../../roadmap.org::*0134. Gas Station][0134. Gas Station]]
 :END:
 
+There are ~n~ gas stations along a circular route, where the amount of gas at the ~i^{th}~ station is ~gas[i]~.
+
+You have a car with an unlimited gas tank and it costs ~cost[i]~ of gas to travel from the ~i^{th}~ station to its next ~(i + 1)^{th}~ station. You begin the journey with an empty tank at one of the gas stations.
+
+Given two integer arrays ~gas~ and ~cost~, return /the starting gas station's index if you can travel around the circuit once in the clockwise direction, otherwise return/ ~-1~. If there exists a solution, it is *guaranteed* to be *unique*.
+
+*Example 1:*
+
+
+#+begin_src
+Input: gas = [1,2,3,4,5], cost = [3,4,5,1,2]
+Output: 3
+Explanation:
+Start at station 3 (index 3) and fill up with 4 unit of gas. Your tank = 0 + 4 = 4
+Travel to station 4. Your tank = 4 - 1 + 5 = 8
+Travel to station 0. Your tank = 8 - 2 + 1 = 7
+Travel to station 1. Your tank = 7 - 3 + 2 = 6
+Travel to station 2. Your tank = 6 - 4 + 3 = 5
+Travel to station 3. The cost is 5. Your gas is just enough to travel back to station 3.
+Therefore, return 3 as the starting index.
+#+end_src
+
+
+*Example 2:*
+
+
+#+begin_src
+Input: gas = [2,3,4], cost = [3,4,3]
+Output: -1
+Explanation:
+You can't start at station 0 or 1, as there is not enough gas to travel to the next station.
+Let's start at station 2 and fill up with 4 unit of gas. Your tank = 0 + 4 = 4
+Travel to station 0. Your tank = 4 - 3 + 2 = 3
+Travel to station 1. Your tank = 3 - 3 + 3 = 3
+You cannot travel back to station 2, as it requires 4 unit of gas but you only have 3.
+Therefore, you can't travel around the circuit once no matter where you start.
+#+end_src
+
+
+*Constraints:*
+
+ - ~n == gas.length == cost.length~
+
+ - ~1 <= n <= 10^{5}~
+
+ - ~0 <= gas[i], cost[i] <= 10^{4}~
+
+ - The input is generated such that the answer is unique.
+
 ** TODO Approach
 Write your approach here.
 
 ** TODO Python
 #+begin_src python
-
+class Solution:
+    def canCompleteCircuit(self, gas: List[int], cost: List[int]) -> int:
 #+end_src
 
 ** TODO C++
 #+begin_src cpp
-
+class Solution {
+public:
+    int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
+        
+    }
+};
 #+end_src