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871. Minimum Number of Refueling Stops

hard

Given fuel stations along a route, find the minimum stops to reach target. A 1D DP indexed by stops, or a heap-based greedy — both classic.

By Alex Chen, Founder, InterviewChamp.AI · Last verified

Problem

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 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.

Constraints

  • 1 <= target, startFuel <= 10^9
  • 0 <= stations.length <= 500
  • 1 <= positioni < positioni+1 < target
  • 1 <= fueli < 10^9

Examples

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.

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Output

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Hints

Progressive — try the first before opening the next.

Hint 1

1D DP: dp[t] = maximum distance reachable using exactly t stops. Update dp[t+1] = max(dp[t+1], dp[t] + fuel) whenever dp[t] >= station position.

Hint 2

Scan stops 1..n and return the smallest t with dp[t] >= target.

Hint 3

Alternative: max-heap. Drive forward; whenever you can't reach the next station, pop the biggest reachable fuel station and 'use' it.

Solution approach

Reveal approach

1D DP approach: define dp[t] as the maximum distance reachable after making exactly t refueling stops. Initialize dp[0] = startFuel and dp[t] = 0 for t > 0. Iterate stations in order of position. For station (p, f): for t from current_max down to 0, if dp[t] >= p then dp[t + 1] = max(dp[t + 1], dp[t] + f). Reverse iteration prevents reusing the same station twice in one pass. After processing every station scan t from 0 upward and return the smallest t with dp[t] >= target, or -1. O(n^2) time, O(n) space. Equivalent O(n log n) greedy uses a max-heap of skipped stations and pops the largest whenever fuel runs out.

Complexity

Time
O(n^2)
Space
O(n)

Related patterns

  • dynamic-programming
  • heap
  • greedy

Related problems

Asked at

Companies reported asking this problem (sourced from public Glassdoor, Blind, and Levels.fyi interview posts).

  • Amazon
  • Google

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