Minimum Number of Refueling Stops
Problem Description
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
Solve it on LeetCode
Approaches
1. Brute Force Recursive Approach
Intuition:
The brute-force approach involves trying all possible combinations of using the available fuel stops to reach the target. We can implement a recursive function that explores each possible stopping point and decides whether to refuel there or not.
While a possible solution, this approach is highly inefficient and not feasible for large input sizes due to its exponential time complexity.
Code:
- Time Complexity: Exponential due to exploring all combinations: O(2^N), where (N) is the number of stations.
- Space Complexity: Depth of recursion: O(N).
2. Dynamic Programming Approach
Intuition:
We can reduce the time complexity by using dynamic programming. Create an array dp where dp[i] is the maximum distance we can reach with i refueling stops. Our goal is to find the smallest i such that dp[i] >= target.
Code:
- Time Complexity: O(N^2), iterating over stations and stops.
- Space Complexity: O(N) for the dp array.
3. Greedy Approach with Max-Heap
Intuition:
The most optimal solution involves using a max-heap to prioritize the largest fuel available at each decision point. This approach picks the station with the most fuel capacity reachable based on the current fuel status at each step.
Code:
- Time Complexity: O(N log N)), primarily due to heap operations.
- Space Complexity: O(N) for the heap structure.