<p>You are given several <code>boxes</code> with different colors represented by different positive numbers.</p>

<p>You may experience several rounds to remove boxes until there is no box left. Each time you can choose some continuous boxes with the same color (i.e., composed of <code>k</code> boxes, <code>k &gt;= 1</code>), remove them and get <code>k * k</code> points.</p>

<p>Return <em>the maximum points you can get</em>.</p>

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

<pre>
<strong>Input:</strong> boxes = [1,3,2,2,2,3,4,3,1]
<strong>Output:</strong> 23
<strong>Explanation:</strong>
[1, 3, 2, 2, 2, 3, 4, 3, 1] 
----&gt; [1, 3, 3, 4, 3, 1] (3*3=9 points) 
----&gt; [1, 3, 3, 3, 1] (1*1=1 points) 
----&gt; [1, 1] (3*3=9 points) 
----&gt; [] (2*2=4 points)
</pre>

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

<pre>
<strong>Input:</strong> boxes = [1,1,1]
<strong>Output:</strong> 9
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> boxes = [1]
<strong>Output:</strong> 1
</pre>

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

<ul>
	<li><code>1 &lt;= boxes.length &lt;= 100</code></li>
	<li><code>1 &lt;= boxes[i]&nbsp;&lt;= 100</code></li>
</ul>

class Solution:
    def removeBoxes(self, b: List[int]) -> int:
        """
        ok watched larry
        i think his sol is hybrid of takeNoTake+interval

        try implement first

        """
        N=len(b)
        @cache
        #streak not yet count self (or consider as #prevs)
        def dfs(l,r,streak):
            if l==r:
                return (streak+1)**2
            elif l>r:
                return 0

            #end streak, new segment
            res=(streak+1)**2 + dfs(l+1,r,0)
            for j in range(l+1,r+1):
                if b[j]==b[l]:  
                    #option to extend streak  via l-j
                    #=> abandon all grouppings of l-[l+1,j-1]
                    res=max(res,dfs(l+1,j-1,0)+dfs(j,r,streak+1))
            return res
        
        return dfs(0,N-1,0)


79000