# burst-balloons


Try it on <a href='https://leetcode.com/problems/burst-balloons'>leetcode</a>

## Description
<div class="description">
<div><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>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>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;= 500</code></li>
	<li><code>0 &lt;= nums[i] &lt;= 100</code></li>
</ul>
</div>
</div>

## Solution(Python)
```Python
class Solution(object):
    def maxCoins(self, nums):
        nums = [1] + nums + [1]
        dp = [[0] * len(nums) for _ in nums]
        n = len(nums)
        for i in range(n - 3, -1, -1):
            for j in range(i + 2, n):
                dp[i][j] = max(
                    [
                        dp[i][k] + dp[k][j] + nums[i] * nums[k] * nums[j]
                        for k in range(i + 1, j)
                    ]
                )
        return dp[0][n - 1]

```