<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) {
        int n = (int)boxes.size();
        int dp[n][n][n]; memset(dp,0,sizeof(dp));
        for(int j = 0; j < n; j++){
            for(int k = 0; k <= j; k++){
                dp[j][j][k] = (k+1)*(k+1);
            }
        }
        for(int len = 1; len < n; len++){
            for(int j = len; j < n; j++){
                int i = j - len;
                for(int k = 0; k <= i; k++){
                    int res = (k+1)*(k+1) + dp[i+1][j][0];
                    for(int m = i+1; m <= j; m++){
                        if(boxes[m] == boxes[i]){
                            res = max(res, dp[i+1][m-1][0]+dp[m][j][k+1]);
                        }
                    }
                    dp[i][j][k] = res;
                }
            }
        }
        return dp[0][n-1][0];
    }
};
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