312. Burst Balloons

Given n balloons, indexed from 0 to n-1. Each balloon is painted with a number on it represented by array nums. You are asked to burst all the balloons. If the you burst balloon i you will get nums[left] * nums[i] * nums[right] coins. Here left and right are adjacent indices of i. After the burst, the left and right then becomes adjacent.
Find the maximum coins you can collect by bursting the balloons wisely.
Note:
* You may imagine nums[-1] = nums[n] = 1. They are not real therefore you can not burst them.
* 0 ≤ n ≤ 500, 0 ≤ nums[i] ≤ 100
Example:
Input: [3,1,5,8]
Output: 167 
Explanation: nums = [3,1,5,8] --> [3,5,8] -->   [3,8]   -->  [8]  --> []
             coins =  3*1*5      +  3*5*8    +  1*3*8      + 1*8*1   = 167



Dfs + memo - > dp 
https://leetcode.com/problems/burst-balloons/discuss/76228/Share-some-analysis-and-explanations



https://www.youtube.com/watch?v=z3hu2Be92UA&t=499s



class Solution {
    public int maxCoins(int[] array) {
        int n = array.length;
        int[] nums = new int[n + 2];
        for(int i = 0; i < array.length; i++){
            nums[i + 1] = array[i];
        }
        nums[0] = 1;
        nums[nums.length - 1] = 1;
        
        // dp[i][j] = maxCoins(nums[i : j + 1])
        int[][] dp = new int[n + 2][n + 2];
        // iterate thru the length from 1 to n 
        for(int l = 1; l <= n; l++){
            // starting point for the length of l 
            for(int i = 1; i < n - l + 1; i++){
                int j = i + l - 1;
                // iterate thru the middle breaking point 
                for(int k = i; k <= j; k++){
                    dp[i][j] = Math.max(dp[i][j], dp[i][k -1] + dp[k + 1][j] + nums[i -1] * nums[k] * nums[j + 1]);
                }
            }
            
        }
        return dp[1][n];
        
    }
}