<p>You are given <code>n</code> balloons, indexed from <code>0</code> to <code>n - 1</code>. Each balloon is painted with a number on it represented by an array <code>nums</code>. You are asked to burst all the balloons.</p>

<p>If you burst the <code>i<sup>th</sup></code> balloon, you will get <code>nums[i - 1] * nums[i] * nums[i + 1]</code> coins. If <code>i - 1</code> or <code>i + 1</code> goes out of bounds of the array, then treat it as if there is a balloon with a <code>1</code> painted on it.</p>

<p>Return <em>the maximum coins you can collect by bursting the balloons wisely</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,1,5,8]
<strong>Output:</strong> 167
<strong>Explanation:</strong>
nums = [3,1,5,8] --&gt; [3,5,8] --&gt; [3,8] --&gt; [8] --&gt; []
coins =  3*1*5    +   3*5*8   +  1*3*8  + 1*8*1 = 167</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,5]
<strong>Output:</strong> 10
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == nums.length</code></li>
	<li><code>1 &lt;= n &lt;= 300</code></li>
	<li><code>0 &lt;= nums[i] &lt;= 100</code></li>
</ul>

class Solution {
public:
    int maxCoins(vector<int>& nums) {
        nums.insert(nums.begin(), 1);
        nums.push_back(1);

        int n = nums.size();
        vector<vector<int>> dp(n, vector<int>(n, 0));

        // 0, n
        int ret = 0;
        for (int l = 3; l <= n; l++)
        {
            for (int i = 0; i + l - 1 < n; i++)
            {
                int tmp = 0;
                for (int j = i + 1; j < i + l - 1; j++)
                {
                    tmp = max(tmp, dp[i][j] + dp[j][i+l-1] + nums[j] * nums[i] * nums[i+l-1]);
                }

                dp[i][i+l-1] = tmp;
                ret = max(ret, dp[i][i+l-1]);

                cout << "i: " << i << ", i+l-1: " << i+l-1 << ", " << dp[i][i+l-1] << endl;
            }
        }

        return ret;
    }
};
12100