<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:
        # Length of the boxes array.
        num_boxes = len(boxes)
        
        # Initialize the memoization table.
        memo = [[[0] * num_boxes for _ in range(num_boxes)] for _ in range(num_boxes)]
        
        # Start the depth-first search to compute the maximum points.
        return self.dfs(boxes, 0, num_boxes - 1, 0, memo)

    def dfs(self, boxes: List[int], start: int, end: int, extra: int, memo: List[List[List[int]]]) -> int:
        # Base case: if the start index is greater than the end index.
        if start > end:
            return 0
        
        # Base case: if the subarray has only one box.
        if start == end:
            return (extra + 1) * (extra + 1)
        
        # Check if the result is already computed and stored in memo.
        if memo[start][end][extra] != 0:
            return memo[start][end][extra]
        
        # Case where the start and end boxes are the same.
        if boxes[start] == boxes[end]:
            memo[start][end][extra] = self.dfs(boxes, start + 1, end, extra + 1, memo)
        else:
            # Initialize the result for the current subarray.
            memo[start][end][extra] = 0
            
            # Try to find and merge subproblems where boxes[start] is the same as any box in the range [start, end].
            for k in range(end - 1, start - 1, -1):
                if boxes[start] == boxes[k]:
                    memo[start][end][extra] = max(
                        self.dfs(boxes, start, k, 0, memo) + self.dfs(boxes, k + 1, end, extra, memo),
                        memo[start][end][extra]
                    )
        
        # Return the computed result for the current subarray.
        return memo[start][end][extra]
26100