I knew this was a greedy problem pretty quickly but fumbled the implementation for a bit.
Use a greedy strategy with a max-heap to always refuel at the station with the most fuel among those passed when the car runs out of gas. Iterate through stations sorted by position, adding reachable stations' fuel to the heap, and when fuel is insufficient to reach the next station or target, pop the max fuel and increment stops. If the heap is empty and fuel is insufficient, return -1.
Pro tip: Clarify that the greedy choice is optimal because refueling at the station with maximum fuel among reachable ones minimizes the number of stops, and mention that this is a classic problem (LeetCode 871) often asked at Meta.
Restate the problem: car starts at 0 with startFuel, stations have positions and fuel amounts, need minimum refueling stops to reach target. Note that fuel consumption is 1 liter per mile, and you can only refuel at stations you pass.
Use a max-heap (priority queue) to store fuel amounts of stations that have been passed but not yet used. Sort stations by position to process them in order.
Iterate through stations and target as checkpoints. At each checkpoint, if current fuel is insufficient to reach it, repeatedly pop the max fuel from the heap and add to current fuel, incrementing stop count, until fuel is enough or heap is empty. If heap is empty and still insufficient, return -1.
If target is reached, return the number of stops. Consider cases like startFuel already enough (return 0), no stations, or unreachable target. Ensure stations beyond target are ignored.
Time complexity: O(n log n) due to sorting and heap operations. Space complexity: O(n) for the heap. Walk through a small example to verify correctness.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.