<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) {
        memset(t, -1, sizeof(t));
		vector<int> arr = {1};
        for(int x: nums) 
			arr.push_back(x);
        arr.push_back(1);
        n = arr.size();
        return solve(arr, 1, arr.size() - 2);
    }
    int n;
    int t[501][501];
    int solve(vector<int> &nums, int i, int j){
		if(i > j)
			return 0;
        if(i == j){
            int temp = nums[i];
            if(i - 1 >= 0)  
                temp *= nums[i - 1];
            if(i + 1 < n)
                temp *= nums[i + 1];
            return temp;
        }
		if(t[i][j] != -1) 
			return t[i][j];
        int ans = 0; 
        for(int k = i; k <= j; k++){
		    int temp = nums[k];
            if(j + 1 < n)  
                temp *= nums[j + 1];	
            if(i - 1 >= 0) 
                temp *= nums[i - 1];
		
            temp += (solve(nums, i, k - 1) + solve(nums, k + 1, j));
            ans = max(ans, temp);
        }
        return t[i][j] = ans;
    }
};
10700