# Car Fleet

`TreeMap`, `Stack`

Medium

`N` cars are going to the same destination along a one lane road. The destination is `target` miles away.

Each car `i` has a constant speed `speed[i]` (in miles per hour), and initial position`position[i]` miles towards the target along the road.

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 distance between these two cars is ignored - they are assumed to have the same position.

A_car fleet_is some non-empty set of cars driving at the same position and same speed. Note that a single car is also a car fleet.

If a car catches up to a car fleet right at the destination point, it will still be considered as one car fleet.

How many car fleets will arrive at the destination?

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 and 8 become a fleet, meeting each other at 12.
The car starting at 0 doesn't catch up to any other car, so it is a fleet by itself.
The cars starting at 5 and 3 become a fleet, meeting each other at 6.
Note that no other cars meet these fleets before the destination, so the answer is 3.``````

Note:

1. `0 <= N <= 10 ^ 4`

2. `0 < target <= 10 ^ 6`

3. `0 < speed[i] <= 10 ^ 6`

4. `0 <= position[i] < target`

5. All initial positions are different.

## Solution & Analysis

### Sorting + Stack

N辆车沿着一条车道驶向位于target英里之外的共同目的地。每辆车i以恒定的速度speed[i]（英里/小时），从初始位置 position[i]（英里）沿车道驶向目的地。 一辆车永远不会超过前面的另一辆车，但它可以追上去，并与前车以相同的速度紧接着行驶。此时，我们会忽略这两辆车之间的距离，也就是说，它们被假定处于相同的位置。车队是一些由行驶在相同位置、具有相同速度的车组成的非空集合。注意，一辆车也可以是一个车队。即便一辆车在目的地才赶上了一个车队，它们仍然会被视作是同一个车队。求会有多少车队到达目的地？

55 ms, faster than 39.83%

Complexity

• Time: O(n + nlogn)

• Space: O(n)

``````class Solution {
class Car {
int position;
double time;

public Car(int position, double time) {
this.position = position;
this.time = time;
}

@Override
public String toString() {
return this.position + ": " + this.time;
}
}
public int carFleet(int target, int[] position, int[] speed) {
int n = position.length;
Car[] cars = new Car[n];
for (int i = 0; i < position.length; i++) {
cars[i] = new Car(position[i], 1.0 * (target - position[i]) / speed[i]);
}

Arrays.sort(cars, (c1, c2) -> Integer.compare(c1.position, c2.position));
Deque<Car> stack = new ArrayDeque<>();

for (Car c: cars) {
while (!stack.isEmpty() && stack.peek().time <= c.time) {
stack.pop();
}
stack.push(c);
}
return stack.size();
}
}``````

### TreeMap

Calculate time needed to arrive the target, sort by the start position. Loop on each car from the end to the beginning.`cur`recorde the current biggest time (the slowest). If another car needs less or equal time than`cur`, it can catch up this car. Otherwise it will become the new slowest car, that is new lead of a car fleet.

Time Complexity: O(NlogN)

Space Complexity

O(N)

``````    public int carFleet(int target, int[] pos, int[] speed) {
TreeMap<Integer, Double> m = new TreeMap<>();
for (int i = 0; i < pos.length; ++i) m.put(-pos[i], (double)(target - pos[i]) / speed[i]);
int res = 0; double cur = 0;
for (double time : m.values()) {
if (time > cur) {
cur = time;
res++;
}
}
return res;
}``````

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