<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:
    def maxCoins(self, nums: List[int]) -> int:
      l = len(nums)
      dp = []
      for i in range(l + 2):
        temp = [0] * (l + 2)
        dp.append(temp)

      nums.append(1)
      nums = [1] + nums

      i = l
      while i >= 1:
        for j in range(l + 1):
          if i > j:
            continue
          res = -1
          for k in range(i, j + 1):
            coins = nums[i - 1] * nums[k] * nums[j + 1] + dp[i][k - 1] + dp[k + 1][j]
            res = max(coins, res)
          dp[i][j] = res
        i -= 1
      return dp[1][l]
      # def recurse(i, j, nums, dp):
      #   if i > j:
      #     return 0
      #   if dp[i][j] != -1:
      #     return dp[i][j]
      #   res = -1
      #   for k in range(i, j + 1):
      #     coins = nums[i - 1] * nums[k] * nums[j + 1] + recurse(i, k - 1, nums, dp) + recurse(k + 1, j, nums, dp)
      #     res = max(res, coins)
      #   dp[i][j] = res
      #   return dp[i][j]
      # return recurse(1, l, nums, dp)

18113