<p>There is a <strong>3 lane road</strong> of length <code>n</code> that consists of <code>n + 1</code> <strong>points</strong> labeled from <code>0</code> to <code>n</code>. A frog <strong>starts</strong> at point <code>0</code> in the <strong>second </strong>lane<strong> </strong>and wants to jump to point <code>n</code>. However, there could be obstacles along the way.</p>

<p>You are given an array <code>obstacles</code> of length <code>n + 1</code> where each <code>obstacles[i]</code> (<strong>ranging from 0 to 3</strong>) describes an obstacle on the lane <code>obstacles[i]</code> at point <code>i</code>. If <code>obstacles[i] == 0</code>, there are no obstacles at point <code>i</code>. There will be <strong>at most one</strong> obstacle in the 3 lanes at each point.</p>

<ul>
	<li>For example, if <code>obstacles[2] == 1</code>, then there is an obstacle on lane 1 at point 2.</li>
</ul>

<p>The frog can only travel from point <code>i</code> to point <code>i + 1</code> on the same lane if there is not an obstacle on the lane at point <code>i + 1</code>. To avoid obstacles, the frog can also perform a <strong>side jump</strong> to jump to <strong>another</strong> lane (even if they are not adjacent) at the <strong>same</strong> point if there is no obstacle on the new lane.</p>

<ul>
	<li>For example, the frog can jump from lane 3 at point 3 to lane 1 at point 3.</li>
</ul>

<p>Return<em> the <strong>minimum number of side jumps</strong> the frog needs to reach <strong>any lane</strong> at point n starting from lane <code>2</code> at point 0.</em></p>

<p><strong>Note:</strong> There will be no obstacles on points <code>0</code> and <code>n</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/03/25/ic234-q3-ex1.png" style="width: 500px; height: 244px;" />
<pre>
<strong>Input:</strong> obstacles = [0,1,2,3,0]
<strong>Output:</strong> 2 
<strong>Explanation:</strong> The optimal solution is shown by the arrows above. There are 2 side jumps (red arrows).
Note that the frog can jump over obstacles only when making side jumps (as shown at point 2).
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/03/25/ic234-q3-ex2.png" style="width: 500px; height: 196px;" />
<pre>
<strong>Input:</strong> obstacles = [0,1,1,3,3,0]
<strong>Output:</strong> 0
<strong>Explanation:</strong> There are no obstacles on lane 2. No side jumps are required.
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/03/25/ic234-q3-ex3.png" style="width: 500px; height: 196px;" />
<pre>
<strong>Input:</strong> obstacles = [0,2,1,0,3,0]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The optimal solution is shown by the arrows above. There are 2 side jumps.
</pre>

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

<ul>
	<li><code>obstacles.length == n + 1</code></li>
	<li><code>1 &lt;= n &lt;= 5 * 10<sup>5</sup></code></li>
	<li><code>0 &lt;= obstacles[i] &lt;= 3</code></li>
	<li><code>obstacles[0] == obstacles[n] == 0</code></li>
</ul>

class Solution:
    def minSideJumps(self, obstacles: List[int]) -> int:
        rocks=[None,[],[],[]]
        for i in range(len(obstacles)-1,-1,-1):
            if obstacles[i]!=0:
                rocks[obstacles[i]].append(i)
        pos=[0,2]
        jumps=0
        while rocks[pos[1]]:
            pos[0]=rocks[pos[1]][-1]-1
            jump_to=None
            for i in range(1,4):
                while rocks[i] and rocks[i][-1]<pos[0]:
                    rocks[i].pop()
                if i!=pos[1]:
                    if not rocks[i]:
                        jump_to=i
                        break
                    if jump_to is None or rocks[i][-1]>rocks[jump_to][-1]:
                        jump_to=i
            pos[1]=jump_to
            jumps+=1
        return jumps
                    
                
        
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