Burst Balloons
<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>
Array; Dynamic Programming
class Solution {
public:
    int maxCoins(vector<int>& nums) {
        int n = nums.size();
        vector<int> arr(n + 2, 1);
        for (int i = 0; i < n; ++i) {
            arr[i + 1] = nums[i];
        }

        vector<vector<int>> f(n + 2, vector<int>(n + 2, 0));
        for (int i = n - 1; i >= 0; --i) {
            for (int j = i + 2; j <= n + 1; ++j) {
                for (int k = i + 1; k < j; ++k) {
                    f[i][j] = max(f[i][j], f[i][k] + f[k][j] + arr[i] * arr[k] * arr[j]);
                }
            }
        }
        return f[0][n + 1];
    }
};