Leetcode #871: Minimum Number of Refueling Stops
In this guide, we solve Leetcode #871 Minimum Number of Refueling Stops in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Greedy, Array, Dynamic Programming, Heap (Priority Queue)
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: target = 1, startFuel = 1, stations = []
Output: 0
Explanation: We can reach the target without refueling.
Python Solution
class Solution:
def minRefuelStops(
self, target: int, startFuel: int, stations: List[List[int]]
) -> int:
pq = []
ans = pre = 0
stations.append([target, 0])
for pos, fuel in stations:
dist = pos - pre
startFuel -= dist
while startFuel < 0 and pq:
startFuel -= heappop(pq)
ans += 1
if startFuel < 0:
return -1
heappush(pq, -fuel)
pre = pos
return ans
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.