My first instinct was DP and I started going down that road before catching myself.
Model the problem as a greedy algorithm: treat each station as a potential refueling point and always refuel at the station with the most fuel among those reachable with the current fuel. Use a max-heap to efficiently select the best station, and count stops until the target is reached or no stations are reachable.
Pro tip: Clarify that the greedy choice is optimal because refueling at the station with maximum fuel maximizes future reach without increasing the number of stops. Also, mention that if the target is unreachable even after considering all stations, return -1.
Restate the problem: given start fuel, target distance, and stations (position, fuel), find the minimum stops to reach the target. Note that fuel is consumed at 1 unit per distance, and you can only refuel at stations you pass.
Use a max-heap to store fuel amounts of reachable stations. Sort stations by position to process them in order as you travel.
Iterate through stations, adding their fuel to the heap when they become reachable. When fuel runs out before the next station or target, pop the max fuel from the heap, add it to current fuel, and increment the stop count.
If the heap is empty and the target is not reachable, return -1. If the target is reached, return the stop count. Ensure stations beyond the target are ignored.
Time complexity is O(n log n) due to sorting and heap operations. Space complexity is 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.