853. Car Fleet

Question

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 array 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 (i.e., 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.

Return the number of car fleets that will arrive at the destination.

Example 1:

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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 car starting at 0 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.
Note that no other cars meet these fleets before the destination, so the answer is 3.

Solution

Firstly, this problem give me some positions. These position a unsorted. So it is obvious that the first step should be sort these cars by position.

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int[][] cars=new int[position.length][2];
    for(int i=0;i<position.length;i++){
        cars[i][0]=position[i];
        cars[i][1]=speed[i];
    }

    Arrays.sort(cars,(a,b)->(a[0]-b[0]));

And then, after the sort, we can fint that when there are two cars(first are is in i, second is in j, i<j). There are two situations:

  1. First car need n time unit to array target and second car need m time unit to array target. If n<=m, these two cars will always become a car fleet.
  2. First car need n time unit to array target and second car need m time unit to array target. If n>m, these two cars will not become a car fleet.

So, we can use a stack to record car fleets. If a new car’s array time is small than peek in stack, we merge this car into peek fleet. On the contrary, the car can make up a new fleet. I just push this car into stack;

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class Solution {
    public int carFleet(int target, int[] position, int[] speed) {
       
        int[][] cars=new int[position.length][2];
        for(int i=0;i<position.length;i++){
            cars[i][0]=position[i];
            cars[i][1]=speed[i];
        }

        Arrays.sort(cars,(a,b)->(a[0]-b[0]));

        List<int[]> stack=new LinkedList<>();

        for(int i=0;i<cars.length;i++){
            if(stack.size()==0){
                stack.add(cars[i]);
                continue;
            }
            double time=((target-cars[i][0])*1.0) / (cars[i][1]*1.0);
            // stack.get(stack.size()-1)[1]
            while(stack.size()!=0 && ((target-stack.get(stack.size()-1)[0])*1.0) / (stack.get(stack.size()-1)[1]*1.0) <= time){
                stack.remove(stack.size()-1);
            }
            stack.add(cars[i]);
        }

        return stack.size();
    }
}