853. Car Fleet #

Problem #

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.

Problem Summary #

N cars drive along a single lane toward a common destination located target miles away. Each car i travels at a constant speed speed[i] (miles per hour), starting from initial position position[i] (miles) along the road toward the destination.

A car can never pass another car ahead of it, but it can catch up to it and then drive bumper to bumper at the same speed. At this point, we ignore the distance between the two cars; in other words, they are assumed to be at the same position. A car fleet is a non-empty set of cars driving at the same position and with the same speed. Note that a single car can also be a car fleet. Even if a car catches up to a car fleet only at the destination, they are still considered the same car fleet.

How many car fleets will arrive at the destination in the end?

Solution Approach #

• Calculate the time for each car to reach the destination based on its distance from the destination and speed, then sort in descending order of distance (a larger position means it is closer to the destination)
• Scan the sorted array from front to back. Once the time is greater than the time of the previous car, a new car fleet is formed, so increment the final count.

Code #

package leetcode

import (
	"sort"
)

type car struct {
	time     float64
	position int
}

// ByPosition define
type ByPosition []car

func (a ByPosition) Len() int           { return len(a) }
func (a ByPosition) Swap(i, j int)      { a[i], a[j] = a[j], a[i] }
func (a ByPosition) Less(i, j int) bool { return a[i].position > a[j].position }

func carFleet(target int, position []int, speed []int) int {
	n := len(position)
	if n <= 1 {
		return n
	}
	cars := make([]car, n)
	for i := 0; i < n; i++ {
		cars[i] = car{float64(target-position[i]) / float64(speed[i]), position[i]}
	}
	sort.Sort(ByPosition(cars))
	fleet, lastTime := 0, 0.0
	for i := 0; i < len(cars); i++ {
		if cars[i].time > lastTime {
			lastTime = cars[i].time
			fleet++
		}
	}
	return fleet
}