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Java-based LeetCode algorithm problem solutions, regularly updated
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package g0801_0900.s0871_minimum_number_of_refueling_stops;

// #Hard #Array #Dynamic_Programming #Greedy #Heap_Priority_Queue
// #2022_03_28_Time_3_ms_(88.52%)_Space_48.8_MB_(60.04%)

import java.util.PriorityQueue;

/**
 * 871 - Minimum Number of Refueling Stops\.
 *
 * Hard
 *
 * A car travels from a starting position to a destination which is `target` miles east of the starting position.
 *
 * There are gas stations along the way. The gas stations are represented as an array `stations` where stations[i] = [position_i, fuel_i] indicates that the i^th gas station is position_i miles east of the starting position and has fuel_i liters of gas.
 *
 * The car starts with an infinite tank of gas, which initially has `startFuel` liters of fuel in it. It uses one liter of gas per one mile that it drives. When the car reaches a gas station, it may stop and refuel, transferring all the gas from the station into the car.
 *
 * Return _the minimum number of refueling stops the car must make in order to reach its destination_. If it cannot reach the destination, return `-1`.
 *
 * Note that if the car reaches a gas station with `0` fuel left, the car can still refuel there. If the car reaches the destination with `0` fuel left, it is still considered to have arrived.
 *
 * **Example 1:**
 *
 * **Input:** target = 1, startFuel = 1, stations = []
 *
 * **Output:** 0
 *
 * **Explanation:** We can reach the target without refueling.
 *
 * **Example 2:**
 *
 * **Input:** target = 100, startFuel = 1, stations = \[\[10,100]]
 *
 * **Output:** -1
 *
 * **Explanation:** We can not reach the target (or even the first gas station).
 *
 * **Example 3:**
 *
 * **Input:** target = 100, startFuel = 10, stations = \[\[10,60],[20,30],[30,30],[60,40]]
 *
 * **Output:** 2
 *
 * **Explanation:** We start with 10 liters of fuel. We drive to position 10, expending 10 liters of fuel.
 *
 * We refuel from 0 liters to 60 liters of gas.
 *
 * Then, we drive from position 10 to position 60 (expending 50 liters of fuel),
 *
 * and refuel from 10 liters to 50 liters of gas. We then drive to and reach the target.
 *
 * We made 2 refueling stops along the way, so we return 2.
 *
 * **Constraints:**
 *
 * *   1 <= target, startFuel <= 10⁹
 * *   `0 <= stations.length <= 500`
 * *   0 <= position_i <= position_i+1 < target
 * *   1 <= fuel_i < 10⁹
**/
public class Solution {
    public int minRefuelStops(int target, int startFuel, int[][] stations) {
        if (startFuel >= target) {
            return 0;
        } else if (stations == null || stations.length == 0) {
            return -1;
        }
        PriorityQueue pq = new PriorityQueue<>((a, b) -> b[1] - a[1]);
        int start = 0;
        int end = stations.length;
        int currentFuel = startFuel;
        int stops = 0;
        while (currentFuel < target) {
            while (start < end && currentFuel >= stations[start][0]) {
                pq.add(stations[start++]);
            }
            if (pq.isEmpty()) {
                return -1;
            }
            int[] current = pq.poll();
            currentFuel += current[1];
            stops++;
        }
        return stops;
    }
}