<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 __init__(self):
        self.dp = {}
        self.obstacles = []
        self.n = 0
        self.maxJumps = 10**10 

    def memo(self, index, lane):
        if (index, lane) in self.dp:
            return self.dp[(index, lane)]
        if index == self.n:
            return 0
        if self.obstacles[index] == lane:
            return self.maxJumps

        self.dp[(index, lane)] = self.maxJumps

        self.dp[(index, lane)] = min(self.dp[(index, lane)], self.memo(index + 1, lane))

        for nextLane in [1, 2, 3]:
            if nextLane != lane and self.obstacles[index] != nextLane:
                self.dp[(index, lane)] = min(self.dp[(index, lane)], 1 + self.memo(index + 1, nextLane))

        return self.dp[(index, lane)]

    def minSideJumps(self, obstacles: list[int]) -> int:
        self.obstacles = obstacles
        self.n = len(obstacles) - 1
        return self.memo(0, 2)  
    # def minSideJumps(self, obstacles: List[int]) -> int:
    #     setrecursionlimit(1000000000)
    #     @cache
    #     def solve(i, j):
    #         if j == n - 1:
    #             return 0
    #         if obstacles[j + 1] == i:
    #             ans = maxsize
    #             for k in (1, 2, 3):
    #                 if k == i or obstacles[j] == k:
    #                     continue
    #                 ans = min(ans, 1 + solve(k, j + 1))
    #             return ans
    #         else:
    #             return solve(i, j + 1)

    #     n = len(obstacles) 
    #     return solve(2, 0)

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