Introduction

In this blog post, we will explore how to solve the LeetCode problem "871" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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] = [positioni, fueli] indicates that the ith gas station is positioni miles east of the starting position and has fueli 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 1 <= positioni < positioni+1 < target 1 <= fueli < 109

Explanation

• Greedy Approach: Iterate through the gas stations, and at each point, if we cannot reach the next station, refuel at the gas station with the most fuel among the stations we've passed but haven't refueled at yet.
  • Priority Queue: Use a max-heap (priority queue) to keep track of the fuel available at the stations we've passed.
  • Handle Edge Cases: If we run out of fuel before reaching the next station or the target, return -1.

• Runtime Complexity: O(N log N), where N is the number of gas stations. This is due to the heap operations.
• Storage Complexity: O(N), where N is the number of gas stations, for the heap.

Code

    import heapq

def minRefuelStops(target: int, startFuel: int, stations: list[list[int]]) -> int:
    """
    Calculates the minimum number of refueling stops required to reach the target.

    Args:
        target: The target distance.
        startFuel: The initial fuel in the car.
        stations: A list of gas stations, where each station is a list [position, fuel].

    Returns:
        The minimum number of refueling stops, or -1 if the target is unreachable.
    """

    stations.append([target, 0])  # Add target as a station with 0 fuel for easier handling
    stops = 0
    current_fuel = startFuel
    pq = []  # Max-heap to store fuel from visited stations
    prev_position = 0

    for position, fuel in stations:
        current_fuel -= (position - prev_position)
        while pq and current_fuel < 0:
            current_fuel += -heapq.heappop(pq)  # Refuel from the station with the most fuel
            stops += 1
        if current_fuel < 0:
            return -1  # Cannot reach the next station or target

        heapq.heappush(pq, -fuel)  # Add fuel to the max-heap
        prev_position = position

    return stops