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1776. Car Fleet II
Description
There are n
cars traveling at different speeds in the same direction along a onelane road. You are given an array cars
of length n
, where cars[i] = [position_{i}, speed_{i}]
represents:
position_{i}
is the distance between thei^{th}
car and the beginning of the road in meters. It is guaranteed thatposition_{i} < position_{i+1}
.speed_{i}
is the initial speed of thei^{th}
car in meters per second.
For simplicity, cars can be considered as points moving along the number line. Two cars collide when they occupy the same position. Once a car collides with another car, they unite and form a single car fleet. The cars in the formed fleet will have the same position and the same speed, which is the initial speed of the slowest car in the fleet.
Return an array answer
, where answer[i]
is the time, in seconds, at which the i^{th}
car collides with the next car, or 1
if the car does not collide with the next car. Answers within 10^{5}
of the actual answers are accepted.
Example 1:
Input: cars = [[1,2],[2,1],[4,3],[7,2]] Output: [1.00000,1.00000,3.00000,1.00000] Explanation: After exactly one second, the first car will collide with the second car, and form a car fleet with speed 1 m/s. After exactly 3 seconds, the third car will collide with the fourth car, and form a car fleet with speed 2 m/s.
Example 2:
Input: cars = [[3,4],[5,4],[6,3],[9,1]] Output: [2.00000,1.00000,1.50000,1.00000]
Constraints:
1 <= cars.length <= 10^{5}
1 <= position_{i}, speed_{i} <= 10^{6}
position_{i} < position_{i+1}
Solutions

class Solution { public double[] getCollisionTimes(int[][] cars) { int n = cars.length; double[] ans = new double[n]; Arrays.fill(ans, 1.0); Deque<Integer> stk = new ArrayDeque<>(); for (int i = n  1; i >= 0; i) { while (!stk.isEmpty()) { int j = stk.peek(); if (cars[i][1] > cars[j][1]) { double t = (cars[j][0]  cars[i][0]) * 1.0 / (cars[i][1]  cars[j][1]); if (ans[j] < 0  t <= ans[j]) { ans[i] = t; break; } } stk.pop(); } stk.push(i); } return ans; } }

class Solution { public: vector<double> getCollisionTimes(vector<vector<int>>& cars) { int n = cars.size(); vector<double> ans(n, 1.0); stack<int> stk; for (int i = n  1; ~i; i) { while (stk.size()) { int j = stk.top(); if (cars[i][1] > cars[j][1]) { double t = (cars[j][0]  cars[i][0]) * 1.0 / (cars[i][1]  cars[j][1]); if (ans[j] < 0  t <= ans[j]) { ans[i] = t; break; } } stk.pop(); } stk.push(i); } return ans; } };

class Solution: def getCollisionTimes(self, cars: List[List[int]]) > List[float]: stk = [] n = len(cars) ans = [1] * n for i in range(n  1, 1, 1): while stk: j = stk[1] if cars[i][1] > cars[j][1]: t = (cars[j][0]  cars[i][0]) / (cars[i][1]  cars[j][1]) if ans[j] == 1 or t <= ans[j]: ans[i] = t break stk.pop() stk.append(i) return ans

func getCollisionTimes(cars [][]int) []float64 { n := len(cars) ans := make([]float64, n) for i := range ans { ans[i] = 1.0 } stk := []int{} for i := n  1; i >= 0; i { for len(stk) > 0 { j := stk[len(stk)1] if cars[i][1] > cars[j][1] { t := float64(cars[j][0]cars[i][0]) / float64(cars[i][1]cars[j][1]) if ans[j] < 0  t <= ans[j] { ans[i] = t break } } stk = stk[:len(stk)1] } stk = append(stk, i) } return ans }