<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 {
private:
int f(int i,int j,vector<int>v,vector<vector<int>>&dp){
    if(i>j) return 0;
    if(dp[i][j]!=-1) return dp[i][j];
    int mini=INT_MIN;
    for(int k=i;k<=j;k++){
        int coins=(v[i-1]*v[k]*v[j+1])+f(i,k-1,v,dp)+f(k+1,j,v,dp);
        mini=max(mini,coins);
    }
    return dp[i][j]=mini;
}
public:
    int maxCoins(vector<int>& v) {
        int n=v.size();
        v.push_back(1);
        v.insert(v.begin(),1);
        // vector<vector<int>>dp(n+1,vector<int>(n+1,-1));
        // return f(1,n,v,dp);

        // tabulation
        vector<vector<int>>dp(n+2,vector<int>(n+2,0));
        for(int i=n;i>=1;i--){
            for(int j=1;j<=n;j++){
                if(i>j) continue;
                int maxi=INT_MIN;
                for(int k=i;k<=j;k++){
                    int coins=(v[i-1]*v[k]*v[j+1])+dp[i][k-1]+dp[k+1][j];
                    maxi=max(maxi,coins);
                }
                dp[i][j]=maxi;
            }
        }
        return dp[1][n];
    }
};
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