<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 topDown(vector<vector<vector<int>>>& DP,vector<pair<int,int>>& G,int i, int j, int k){
        if(i > j || i == G.size())return 0;
        if(DP[i][j][k] != -1)return DP[i][j][k];
        DP[i][j][k]=0;
        DP[i][j][k] = (G[i].second + k) * (G[i].second + k) + topDown(DP,G,i+1,j,0);
        for(int g=i+1;g<=j;g++){
            if(G[g].first == G[i].first)DP[i][j][k] = max(DP[i][j][k],topDown(DP,G,i+1,g-1,0)+topDown(DP,G,g,j,k+G[i].second));
        }
        return DP[i][j][k];
    }

    int removeBoxes(vector<int>& boxes) {
    //BruteForce -- Using recursion -- NOT DONE -- DO IT.
    //Efficient Approach -- Using 3D DP -- TC:O(n*n*n*n) && SC:O(n*n*n).
        int n=boxes.size();
        vector<pair<int,int>> G;
        for(int i=0;i<n;i++){
            int j=i;
            while(i+1 < n && boxes[i+1] == boxes[j])i++;
            G.push_back({boxes[j],i-j+1});
        }

        int m=G.size();
        vector<vector<vector<int>>> DP(m+1,vector<vector<int>>(m+1,vector<int>(n+1,-1)));
        return topDown(DP,G,0,m-1,0);
    }
};
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