<p>There are <code>n</code> cars on an infinitely long road. The cars are numbered from <code>0</code> to <code>n - 1</code> from left to right and each car is present at a <strong>unique</strong> point.</p>

<p>You are given a <strong>0-indexed</strong> string <code>directions</code> of length <code>n</code>. <code>directions[i]</code> can be either <code>&#39;L&#39;</code>, <code>&#39;R&#39;</code>, or <code>&#39;S&#39;</code> denoting whether the <code>i<sup>th</sup></code> car is moving towards the <strong>left</strong>, towards the <strong>right</strong>, or <strong>staying</strong> at its current point respectively. Each moving car has the <strong>same speed</strong>.</p>

<p>The number of collisions can be calculated as follows:</p>

<ul>
	<li>When two cars moving in <strong>opposite</strong> directions collide with each other, the number of collisions increases by <code>2</code>.</li>
	<li>When a moving car collides with a stationary car, the number of collisions increases by <code>1</code>.</li>
</ul>

<p>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.</p>

<p>Return <em>the <strong>total number of collisions</strong> that will happen on the road</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> directions = &quot;RLRSLL&quot;
<strong>Output:</strong> 5
<strong>Explanation:</strong>
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. 
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> directions = &quot;LLRR&quot;
<strong>Output:</strong> 0
<strong>Explanation:</strong>
No cars will collide with each other. Thus, the total number of collisions that will happen on the road is 0.</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= directions.length &lt;= 10<sup>5</sup></code></li>
	<li><code>directions[i]</code> is either <code>&#39;L&#39;</code>, <code>&#39;R&#39;</code>, or <code>&#39;S&#39;</code>.</li>
</ul>

class Solution:
    def countCollisions(self, directions: str) -> int:
        count = 0
        # would be faster if we did not turn it into the string and instead just make it prev and current. Sincw we only need to know the now and before
        queue = list(directions)[::-1]
        empty_list = []
        crash = 0
        while queue:
            if not empty_list:
                empty_list.append(queue.pop())
                continue
            top = empty_list[-1]
            curr = queue.pop()
            if top == 'R' and curr == 'L':
                crash +=2
                empty_list.pop()
                queue.append('S')
            elif top == 'S' and curr == "L":
                crash +=1
                empty_list.append('S')
            elif top == 'R' and curr == "S":
                crash +=1
                empty_list.pop()
                queue.append('S')
            else:
                empty_list.append(curr)
        return crash



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