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Formatted question description: https://leetcode.ca/all/853.html
853. Car Fleet
Level
Medium
Description
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 nonempty 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:
0 <= N <= 10 ^ 4
0 < target <= 10 ^ 6
0 < speed[i] <= 10 ^ 6
0 <= position[i] < target
 All initial positions are different.
Solution
For each car, calculate the time to reach the destination. Create an array that stores each ear’s initial poisition and time to reach the destination, and sort the array according to the cars’ initial positions in ascending order. Then loop over the array in reversing order. For each car, if it can catch up to the car before it, then assign the current element in the array to the previous element in the array. Otherwise, increase the number of car fleets by 1. Finally, return the number of car fleets.

class Solution { public int carFleet(int target, int[] position, int[] speed) { int length = position.length; if (length == 0) return 0; Struct[] array = new Struct[length]; for (int i = 0; i < length; i++) array[i] = new Struct(position[i], 1.0 * (target  position[i]) / speed[i]); Arrays.sort(array); int fleets = 1; int index = length  1; while (index > 0) { if (array[index].time < array[index  1].time) fleets++; else array[index  1] = array[index]; index; } return fleets; } } class Struct implements Comparable<Struct> { int position; double time; public Struct(int position, double time) { this.position = position; this.time = time; } public int compareTo(Struct struct2) { return this.position  struct2.position; } }

// OJ: https://leetcode.com/problems/carfleet/ // Time: O(NlogN) // Space: O(N) class Solution { public: int carFleet(int target, vector<int>& P, vector<int>& S) { int N = P.size(), ans = 0; vector<int> id(N); iota(begin(id), end(id), 0); sort(begin(id), end(id), [&](int a, int b) { return P[a] < P[b]; }); for (int i = N  1; i >= 0; i) { float time = (float)(target  P[id[i]]) / S[id[i]]; while (i  1 >= 0 && (float)(target  P[id[i  1]]) / S[id[i  1]] <= time) i; ++ans; } return ans; } };

class Solution: def carFleet(self, target, position, speed): """ :type target: int :type position: List[int] :type speed: List[int] :rtype: int """ car = [(pos, spe) for pos, spe in zip(position, speed)] car.sort(reverse=True) time = [(target  pos) / spe for pos, spe in car] ls = [] for i in time: if not ls: ls.append(i) else: if i > ls[1]: ls.append(i) return len(ls) ############ 3 class Solution: def carFleet(self, target, position, speed): """ :type target: int :type position: List[int] :type speed: List[int] :rtype: int """ cars = [(pos, spe) for pos, spe in zip(position, speed)] sorted_cars = sorted(cars) times = [(target  pos) / spe for pos, spe in sorted_cars] stack = [] for time in times[::1]: if not stack: stack.append(time) else: if time > stack[1]: stack.append(time) return len(stack)