<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, boxes: List[int]) -> int:
        # Preprocess the boxes array to condense contiguous blocks of the same color
        condensed = []
        count = []
        n = len(boxes)

        i = 0
        while i < n:
            j = i
            while j < n and boxes[j] == boxes[i]:
                j += 1
            condensed.append(boxes[i])
            count.append(j - i)
            i = j

        memo = {}

        def helper(l: int, r: int, k: int) -> int:
            if l > r:
                return 0
            if (l, r, k) in memo:
                return memo[(l, r, k)]
            
            # Base case
            result = helper(l, r - 1, 0) + (count[r] + k) ** 2
            
            for i in range(l, r):
                if condensed[i] == condensed[r]:
                    result = max(result, 
                                 helper(l, i, count[r] + k) + helper(i + 1, r - 1, 0))
            
            memo[(l, r, k)] = result
            return result
        
        return helper(0, len(condensed) - 1, 0)
26800