2211. Count Collisions on a Road

Medium

459

197

There are `n`

cars on an infinitely long road. The cars are numbered from `0`

to `n - 1`

from left to right and each car is present at a **unique** point.

You are given a **0-indexed** string `directions`

of length `n`

. `directions[i]`

can be either `'L'`

, `'R'`

, or `'S'`

denoting whether the `i`

car is moving towards the ^{th}**left**, towards the **right**, or **staying** at its current point respectively. Each moving car has the **same speed**.

The number of collisions can be calculated as follows:

- When two cars moving in
**opposite**directions collide with each other, the number of collisions increases by`2`

. - When a moving car collides with a stationary car, the number of collisions increases by
`1`

.

After a collision, the cars involved can no longer move and will stay at the point where they collided. Other than that, cars cannot change their state or direction of motion.

Return *the total number of collisions that will happen on the road*.

**Example 1:**

Input:directions = "RLRSLL"Output:5Explanation:The collisions that will happen on the road are: - Cars 0 and 1 will collide with each other. Since they are moving in opposite directions, the number of collisions becomes 0 + 2 = 2. - Cars 2 and 3 will collide with each other. Since car 3 is stationary, the number of collisions becomes 2 + 1 = 3. - Cars 3 and 4 will collide with each other. Since car 3 is stationary, the number of collisions becomes 3 + 1 = 4. - Cars 4 and 5 will collide with each other. After car 4 collides with car 3, it will stay at the point of collision and get hit by car 5. The number of collisions becomes 4 + 1 = 5. Thus, the total number of collisions that will happen on the road is 5.

**Example 2:**

Input:directions = "LLRR"Output:0Explanation:No cars will collide with each other. Thus, the total number of collisions that will happen on the road is 0.

**Constraints:**

`1 <= directions.length <= 10`

^{5}`directions[i]`

is either`'L'`

,`'R'`

, or`'S'`

.

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