<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:
    import math
    def minSideJumps(self, obstacles: List[int]) -> int:
        n=len(obstacles)-1
        tab = [[ int(math.pow(2,31)-1) for _ in range(len(obstacles))] for _ in range(4)]
        for l in range(4):
            tab[l][n] = 0
        for i in reversed(range(n)): #position
            for k in range(1,4): #lane
                if(obstacles[i+1]== k):
                    ans = int(math.pow(2,31)-1)
                    for j in range(1,4):
                        if k!=j and obstacles[i]!=j:
                            ans = min(ans, 1 + tab[j][i+1]) #minJumps(n, obs, i, j)
                    tab[k][i] = ans
                else:
                    tab[k][i] = tab[k][i+1]  #minJumps(n, obs, i+1, curr_lane)
        a =  min(tab[2][0], min(1+tab[1][0], 1+tab[3][0]))
        return a
        # mem = [[-1 for _ in range(len(obstacles))] for _ in range(4)]
        # count = 0
        # def minJumps(n,obs, i, curr_lane):
        #     if(i==n):
        #         mem[curr_lane][i]=0
        #         return mem[curr_lane][i]
        #     if mem[curr_lane][i]!=-1:
        #         return mem[curr_lane][i]
        #     if(obs[i+1]==curr_lane):
        #         ans = 99999999999
        #         for j in range(1,4):
        #             if j!=curr_lane and obs[i]!=j:
        #                 ans = min(ans, 1+ minJumps(n, obs, i, j))
        #         mem[curr_lane][i]= ans
        #     else:
        #         mem[curr_lane][i]= minJumps(n, obs, i+1, curr_lane)
        #     return mem[curr_lane][i]
        # a = minJumps(len(obstacles)-1, obstacles, 0, 2)
        # # print(mem)
        # return a
        # def minJumps(n,obs, i, curr_lane):
        #     if(i==n):
        #         return 0
        #     if(obs[i+1]==curr_lane):
        #         ans = 99999999999
        #         for j in range(1,4):
        #             if j!=curr_lane and obs[i]!=j:
        #                 ans = min(ans, 1+minJumps(n, obs, i, j))
        #         return ans
        #     else:
        #         return minJumps(n, obs, i+1, curr_lane)
        # return minJumps(len(obstacles)-1, obstacles, 0, 2)
        
102600