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.

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.
 Input: target = 10, position = [3], speed = [3]
Output: 1
Explanation: There is only one car, hence there is only one fleet. 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 fleet moves at speed 2.
Then, the fleet (speed 2) and the car starting at 4 (speed 1) become one fleet, meeting each other at 6. The fleet moves at speed 1 until it reaches target.
Constraints:

Contents

In this problem statement, it is mentioned that the cars are going in a single lane road, this means that if there is a slower ahead of a faster car, then the faster car will join fleet along with the slower car.

So, in this approach, since we dont know which car the is closest one to the destination, whether it is a faster car or slower car, previous position car can only either join fleet along with the closest car to destination or it will form a new fleet. Inorder to achieve this, we are going to sort the position array along with speed as a pair. Then from each position, we will compute, how much time it would take to reach to the destination using remaining distance to destination / speed, After this step, we will store these time values into a stack, before inserting a new time into the stack, we will check if the top element on the stack if greater than current element, that means current car cannot overtake the one in front of it, though it is going at a faster speed, since it is a single lane road, so this car will join the fleet along with the car already going in front of it. So, we will not add this element into the stack, and continue this for all positions, and at the end number of elements we will have it in stack are the total number of fleets.

Let's look at an example, position = [10,8,0,5,3] and speed = [2,4,1,1,3] and target = 12.

import java.util.Arrays; import java.util.Comparator; import java.util.Stack; import java.util.stream.Stream; public class CarFleet { static int carFleetUsingSorting(int target, int[] position, int[] speed) { int[][] positionAndSpeedPair = new int[position.length][2]; for(int i=0; i<position.length; i++) { // position and speed are same length arrays positionAndSpeedPair[i][0] = position[i]; positionAndSpeedPair[i][1] = speed[i]; } // sort based on position, reverse order since we want to sort based on highest position to the lowest position Arrays.sort(positionAndSpeedPair, Comparator.comparing(o -> o[0], Comparator.reverseOrder())); Stack<Double> stack = new Stack<>(); Stream.of(positionAndSpeedPair).forEach(pair-> { // compute the time it takes to reach destination // distance to destination from current position / speed double timeToReach = (double)(target - pair[0]) / pair[1]; if(stack.isEmpty() || timeToReach > stack.peek()) { stack.push(timeToReach); } }); return stack.size(); } public static void main(String[] args) { System.out.println(carFleetUsingSorting(12, new int[]{10,8,0,5,3}, new int[]{2,4,1,1,3})); } }
Complexity Analysis:

Time complexity: All the above operations runs in O(n log n) time since we are sorting the position array, where n is the length of the position array.
Space complexity: O(n) for storing stack elements into stack.

In this approach, we are going to pre-compute time takes for each position to reach destination and store it in a temporary array time. Then, we will loop through from right to left, because the closest position will reach the destinaiton first and it is going to block other cars before that, either they have to join the fleet along with that or form their own fleet.

Let's look at implementation steps with an example, position = [10,8,0,5,3] and speed = [2,4,1,1,3] and target = 12.

public class CarFleet { static int carFleet(int target, int[] position, int[] speed) { double[] time = new double[target]; // position[i] < target, so length n is good enough for(int i=0; i<position.length; i++) { time[position[i]] = (double)(target - position[i])/ speed[i]; } int result = 0; double prev = 0.0; for(int i=time.length-1; i>=0; i--) { double current = time[i]; if(current > prev) { prev = current; result++; } } return result; } public static void main(String[] args) { System.out.println(carFleet(12, new int[]{10,8,0,5,3}, new int[]{2,4,1,1,3})); } }
Complexity Analysis:

Time complexity: All the above operations runs in O(n) time, where n is the destination.
Space complexity: O(n) for the temporary time array with destination as size.

Above implementations source code can be found at GitHub link for Java code