# 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。

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;
}
``````