<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:
        cache = {}
        N = len(nums)
        nums = [1] + nums + [1]

        def dfs(l, r):
            if l > r:
                return 0
            
            if (l, r) in cache:
                return cache[(l, r)]
            
            cache[(l , r)] = 0

            for i in range(l, r + 1):
                coins = nums[l - 1] * nums[i] * nums[r + 1]
                coins += dfs(l, i - 1) + dfs(i + 1, r)
                cache[(l , r)] = max(cache[(l , r)], coins)

            return cache[(l , r)]
        
        return dfs(1, len(nums) - 2)
            
30949