leetcode/apps/docs/content/(array)/853-car-fleet.mdx

---
title: '853. Car Fleet'
description: There are n cars at given miles away from the starting mile 0, traveling to reach the mile target
sidebar:
  label: 'Car Fleet'
  badge: 'Medium'
---

<Badge variant="accent">Monotonic Stack</Badge>

### Example 1:
- Input: `target = 12, position = [10,8,0,5,3], speed = [2,4,1,1,3]`
- Output: `3`
- Explanation: • The cars starting at `10 (speed 2`) and `8 (speed 4`) become a fleet, meeting each other at `12`. The fleet forms at `target`. • The car starting at `0 (speed 1`) does not catch up to any other car, so it is a fleet by itself. • The cars starting at `5 (speed 1`) and `3 (speed 3`) become a fleet, meeting each other at `6`. The fleet moves at `speed 1` until it reaches `target`.

### Example 2:
- Input: `target = 10, position = [3], speed = [3]`
- Output: `1`
- Explanation: There is only one car, hence there is only one fleet.

### Example 3:
- Input: `target = 100, position = [0,2,4], speed = [4,2,1]`
- Output: `1`
- Explanation: • The cars starting at `0 (speed 4`) and `2 (speed 2`) become a fleet, meeting each other at `4`. The car starting at `4 (speed 1`) travels to `5`. • Then, the fleet at `4 (speed 2`) and the car at `position 5 (speed 1`) become one fleet, meeting each other at `6`. The fleet moves at `speed 1` until it reaches `target`.

### Constraints:

- `n == position.length == speed.length`
- `1 <= n <= 10^5`
- `0 < target <= 10^6`
- `0 <= position[i] < target`
- All the values of `position` are unique.
- `0 < speed[i] <= 10^6`

## Approach

```mermaid
flowchart TD
  S(["carFleet(target, position, speed)"]) --> O["zip and sort by position, descending — nearest the target first"]
  O --> I["stack = [] — arrival time of each fleet's leader"]
  I --> L{"more (pos, spd)?"}
  L -- no --> E(["return len(stack) — one entry per fleet"])
  L -- yes --> T["time = (target - pos) / spd"]
  T --> Q{"stack empty or time > stack[-1]?"}
  Q -- yes --> A["stack.append(time) — slower, so it starts a new fleet"]
  Q -- no --> B["discard — it catches the car ahead and joins that fleet"]
  A --> L
  B --> L
```

## Solution

```py
class Solution:
    def carFleet(self, target: int, position: List[int], speed: List[int]) -> int:
        stack = []
        for pos, spd in sorted(zip(position, speed), reverse=True):
            time = (target - pos) / spd
            if not stack or time > stack[-1]:
                stack.append(time)

        return len(stack)
```

## Explanation

<YouTube id="Pr6T-3yB9RM" title="Car Fleet - Leetcode 853 - Python" />