There are `n`

cars traveling at different speeds in the same direction along a one-lane road. You are given an array `cars`

of length `n`

, where `cars[i] = [position`

represents:_{i}, speed_{i}]

`position`

is the distance between the_{i}`i`

car and the beginning of the road in meters. It is guaranteed that^{th}`position`

._{i}< position_{i+1}`speed`

is the initial speed of the_{i}`i`

car in meters per second.^{th}

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`

car collides with the next car, or ^{th}`-1`

if the car does not collide with the next car. Answers within `10`

of the actual answers are accepted.^{-5}

**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}

**Solution: Monotonic Stack**

Key observation: If my speed is slower than the speed of the previous car, not only mine but also all cars behind me will NEVER be able to catch/collide with the previous car. Such that we can throw it away.

Maintain a stack that stores the indices of cars with increasing speed.

Process car from right to left, for each car, pop the stack (throw the fastest car away) if any of the following conditions hold.

1) speed <= stack.top().speed

2) There are more than one car before me and it takes more than to collide the fastest car than time the fastest took to collide.

Time complexity: O(n)

Space complexity: O(n)

## C++

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// Author: Huahua class Solution { public: vector<double> getCollisionTimes(vector<vector<int>>& cars) { auto collide = [&](int i, int j) -> double { return static_cast<double>(cars[i][0] - cars[j][0]) / (cars[j][1] - cars[i][1]); }; const int n = cars.size(); vector<double> ans(n, -1); stack<int> s; for (int i = n - 1; i >= 0; --i) { while (!s.empty() && (cars[i][1] <= cars[s.top()][1] || (s.size() > 1 && collide(i, s.top()) > ans[s.top()]))) s.pop(); ans[i] = s.empty() ? -1 : collide(i, s.top()); s.push(i); } return ans; } }; |

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