How do you solve Minimum Number of Refueling Stops on LeetCode?

Minimum Number of Refueling Stops (LeetCode #871) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Greedy algorithms can often yield optimal solutions for problems involving choices over time.

What is the time complexity of Minimum Number of Refueling Stops?

The optimal solution for Minimum Number of Refueling Stops runs in O(n log n) time and uses O(n) space. The time complexity is O(n log n) due to the priority queue operations for each station, where n is the number of stations.

What is the brute force solution for Minimum Number of Refueling Stops?

The brute force approach for Minimum Number of Refueling Stops is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach tries every possible combination of refueling stops to see if the destination can be reached. This method is straightforward but inefficient, as it explores all paths without optimization. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Minimum Number of Refueling Stops?

Minimum Number of Refueling Stops uses the following concepts: Array, Dynamic Programming, Greedy, Heap (Priority Queue). The recommended patterns to study are: Greedy, Priority Queue, Dynamic Programming.

What are the interview tips for Minimum Number of Refueling Stops?

Always clarify the problem statement and constraints before diving into coding. Think aloud during the interview to show your thought process and reasoning. Consider edge cases and test your solution against them after implementing.

What are common mistakes when solving Minimum Number of Refueling Stops?

Failing to consider edge cases, such as when the car cannot reach the first station. Not using a priority queue effectively, leading to suboptimal refueling choices.

#871

Minimum Number of Refueling Stops

Hard

ArrayDynamic ProgrammingGreedyHeap (Priority Queue)GreedyPriority QueueDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

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Intuition

Time O(n log n)Space O(n)

The optimal solution uses a greedy approach with a priority queue to always refuel at the station with the most fuel available when needed. This ensures that we minimize the number of stops while maximizing the fuel we can carry forward.

⚙️

Algorithm

4 steps

1Step 1: Initialize a max-heap (priority queue) to store fuel from reachable stations.
2Step 2: Iterate through each station, checking if it can be reached with current fuel.
3Step 3: If reachable, add its fuel to the heap and check if the destination can be reached with current fuel.
4Step 4: If not, refuel from the station with the most fuel from the heap until we can reach the destination or exhaust options.

solution.py15 lines

1import heapq
2
3def minRefuelStops(target, startFuel, stations):
4    max_heap = []
5    stops = 0
6    current_fuel = startFuel
7    stations.append((target, 0))
8    for position, fuel in stations:
9        while current_fuel < position:
10            if not max_heap:
11                return -1
12            current_fuel += -heapq.heappop(max_heap)
13            stops += 1
14        heapq.heappush(max_heap, -fuel)
15    return stops

ℹ

Complexity note: The time complexity is O(n log n) due to the priority queue operations for each station, where n is the number of stations.

1Greedy algorithms can often yield optimal solutions for problems involving choices over time.
2Using a priority queue allows us to efficiently manage the best options available at each step.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.