Car Fleet
Detailed guide and Python implementation for the 'Car Fleet' problem.
1. Узнать
The 'Car Fleet' problem is a key challenge in the Stack section.
This implementation focuses on easy-level logic in Python.
We prioritize technical accuracy and code readability in our provided solutions.
2. Real-World Applications
3. Visual Intuition
Visualizing the logic flow for Car Fleet.
4. Prerequisites
5. Step-by-Step Thinking
1. Understand the problem
Read the problem statement for Car Fleet carefully.
2. Formulate brute force
Draft a simple iterative solution.
3. Identify inefficiency
Look for redundant calculations.
4. Optimize search path
Use hashing or sorting to speed up the process.
5. Final Implementation
Clean up the code for production standards.
Постановка задачи
There are n cars going to the same destination along a one-lane road. The destination is target miles away.
You are given two integer arrays position and speed, both of length n, where position[i] is the position of the ith car and speed[i] is the speed of the ith car (in miles per hour).
A car can never pass another car ahead of it, but it can catch up to it and drive bumper to bumper at the same speed. The faster car will slow down to match the slower car's speed. The distance between these two cars is ignored (they are assumed to be at the same position).
A car fleet is some non-empty set of cars driving at the same position and same speed. A single car is also a car fleet.
Return the number of car fleets that will arrive at the destination.
Write a function carFleet(target: int, position: List[int], speed: List[int]) -> int.
- •n == len(position) == len(speed)
- •1 <= n <= 10^5
- •0 < target <= 10^6
- •0 <= position[i] < target
- •0 < speed[i] <= 10^6
- •All positions are unique
Примеры
target = 12, position = [10, 8, 0, 5, 3], speed = [2, 4, 1, 1, 3]
3
Cars at positions 10 and 8: car at 8 catches car at 10 (both arrive at time 1), forming 1 fleet. Car at 0: arrives at time 12. Car at 5: arrives at time 7. Car at 3: arrives at time 3, catches car at 5 at time 7, but car at 5 arrives at 7 too. Cars at 3 and 5 form a fleet. Total: 3 fleets.
target = 10, position = [3], speed = [3]
1
Only one car, so one fleet.
target = 100, position = [0, 2, 4], speed = [4, 2, 1]
1
All cars eventually form a single fleet.
Need a Hint?
Edge Cases to Watch
- Empty input structures
- Single element inputs
- Large numerical bounds
Готовы решить?
Open the problem in PyRun's browser-based Python editor. Your code runs fully offline — no server required.
Интервью: идеи и вариации
Разбивка анализа сложности
Почему время: Directly evaluates all possibilities.
Почему космос: Uses standard local memory.
Почему время: Optimized paths reduce total operations.
Почему космос: May trade memory for speed.
Оптимизированный код Python для решения
Оптимизированный код Python для решения
def car_fleet_opt(target, position, speed):
pair = [[p, s] for p, s in zip(position, speed)]
stack = []
for p, s in sorted(pair)[::-1]:
stack.append((target - p) / s)
if len(stack) >= 2 and stack[-1] <= stack[-2]:
stack.pop()
return len(stack)Код грубой силы (спойлер защищен)
Код грубой силы (спойлер защищен)
def car_fleet_brute(target, position, speed):
cars = sorted(zip(position, speed), reverse=True)
times = [(target - p) / s for p, s in cars]
fleets = 0
curr_max_time = 0
for t in times:
if t > curr_max_time:
fleets += 1
curr_max_time = t
return fleetsAlgorithm Pattern Checklist
When dealing with Stack data patterns.
- Are constraints clear?
- Is there a linear or logarithmic optimization possible?
Key Revision Notes
Standard Stack problem properties apply.
Связанные вопросы
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