<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 {
 public:
  int removeBoxes(vector<int>& boxes) {
    const int n = boxes.size();
    vector<vector<vector<int>>> mem(n, vector<vector<int>>(n, vector<int>(n)));
    return removeBoxes(boxes, 0, n - 1, 0, mem);
  }

 private:
  // Returns the maximum score of boxes[i..j] if k boxes eqaul to boxes[j].
  int removeBoxes(const vector<int>& boxes, int i, int j, int k,
                  vector<vector<vector<int>>>& mem) {
    if (i > j)
      return 0;
    if (mem[i][j][k] > 0)
      return mem[i][j][k];

    int r = j;
    int sameBoxes = k + 1;
    while (r > 0 && boxes[r - 1] == boxes[r]) {
      --r;
      ++sameBoxes;
    }
    mem[i][j][k] = removeBoxes(boxes, i, r - 1, 0, mem) + sameBoxes * sameBoxes;

    for (int p = i; p < r; ++p)
      if (boxes[p] == boxes[r])
        mem[i][j][k] =
            max(mem[i][j][k], removeBoxes(boxes, i, p, sameBoxes, mem) +
                                  removeBoxes(boxes, p + 1, r - 1, 0, mem));

    return mem[i][j][k];
  }
};
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