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# Minimum Number of Refueling Stops

A car travels from a starting position to a destination which is `target` miles east of the starting position.

Along the way, there are gas stations.  Each `station[i]` represents a gas station that is `station[i][0]` miles east of the starting position, and has `station[i][1]` liters of gas.

The car starts with an infinite tank of gas, which initially has `startFuel` liters of fuel in it.  It uses 1 liter of gas per 1 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.

What is the least 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't 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.
```

**Note:**

1. `1 <= target, startFuel, stations[i][1] <= 10^9`
2. `0 <= stations.length <= 500`
3. `0 < stations[0][0] < stations[1][0] < ... < stations[stations.length-1][0] < target`

```java
class Solution {
    // Priority Queue O(NlogN)
    public int minRefuelStops(int target, int cur, int[][] s) {
        Queue<Integer> pq = new PriorityQueue<>((a, b) -> b - a);
        int i = 0, stops = 0;
        while (cur < target) {
            // Finding all the stations within current range
            // Choose the station with max fuel overall
            while (i < s.length && s[i][0] <= cur)
                pq.add(s[i++][1]);
            if (pq.isEmpty())
                return -1;
            // Remember our PQ is common for multiple loops
            // so lets say we take max from a loop and cannot reach anywhere after that,
            // then we will also take a stop at 2nd highest value and so on until
            // either our PQ becomes empty or our range reaches somewhere
            cur += pq.poll();
            stops++;
        }
        return stops;
    }
}
```


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