Leetcode #2188: Minimum Time to Finish the Race
In this guide, we solve Leetcode #2188 Minimum Time to Finish the Race 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
You are given a 0-indexed 2D integer array tires where tires[i] = [fi, ri] indicates that the ith tire can finish its xth successive lap in fi * ri(x-1) seconds. For example, if fi = 3 and ri = 2, then the tire would finish its 1st lap in 3 seconds, its 2nd lap in 3 * 2 = 6 seconds, its 3rd lap in 3 * 22 = 12 seconds, etc.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Array, Dynamic Programming
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: tires = [[2,3],[3,4]], changeTime = 5, numLaps = 4
Output: 21
Explanation:
Lap 1: Start with tire 0 and finish the lap in 2 seconds.
Lap 2: Continue with tire 0 and finish the lap in 2 * 3 = 6 seconds.
Lap 3: Change tires to a new tire 0 for 5 seconds and then finish the lap in another 2 seconds.
Lap 4: Continue with tire 0 and finish the lap in 2 * 3 = 6 seconds.
Total time = 2 + 6 + 5 + 2 + 6 = 21 seconds.
The minimum time to complete the race is 21 seconds.
Python Solution
class Solution:
def minimumFinishTime(
self, tires: List[List[int]], changeTime: int, numLaps: int
) -> int:
cost = [inf] * 18
for f, r in tires:
i, s, t = 1, 0, f
while t <= changeTime + f:
s += t
cost[i] = min(cost[i], s)
t *= r
i += 1
f = [inf] * (numLaps + 1)
f[0] = -changeTime
for i in range(1, numLaps + 1):
for j in range(1, min(18, i + 1)):
f[i] = min(f[i], f[i - j] + cost[j])
f[i] += changeTime
return f[numLaps]
Complexity
The time complexity is O(n·m) (typical). The space complexity is O(n·m) or optimized.
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.