Minimize Refueling Stops on a One-Direction Journey

### Problem A car starts at coordinate0 with `startFuel` liters and travels east to `target`. It consumes one liter per mile and has an unlimited-capacity tank. Each station `[position,fuel]` offers all its fuel when you choose to stop there. Return the minimum number of refueling stops needed to reach target, or `-1` if impossible. Reaching a station or the destination with exactly zero fuel is allowed. ```cpp class Solution { public: int minRefuelStops(int target, int startFuel, vector<vector<int>>& stations); }; ``` ### Examples **Example 1:** ```text Input: target = 50, startFuel = 10, stations = [[10,20],[25,25]] Output: 2 Explanation: The first stop extends reach to30; the second extends reach to55. One stop cannot reach50. ``` **Example 2:** ```text Input: target = 20, startFuel = 5, stations = [[6,20]] Output: -1 Explanation: The first station is one mile beyond the initial reach. ``` ### Constraints - Both target and the initial fuel are integers in the inclusive interval `[1,1000000000]`. - There may be zero stations; at most500 are supplied. - Station positions are integers at least1, strictly increasing, and strictly less than target. - Each station supplies an integer fuel amount satisfying `1 <= fuel_i < 1000000000`. ### Contract considerations All station positions are strictly before target. Transferring fuel can be modeled retrospectively because every heap station was already reachable. ```hint Read the boundary condition When forced to add one stop, which reachable station gives the greatest possible extension? ``` ### Boundary reminders All station positions are strictly before target. Transferring fuel can be modeled retrospectively because every heap station was already reachable.