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: 
(1) You may imagine nums[-1] = nums[n] = 1. They are not real therefore you can not burst them.
(2) 0 ≤ n ≤ 500, 0 ≤ nums[i] ≤ 100

Example:

Given [3, 1, 5, 8]

Return 167

    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

Code

public class Solution {
    public int maxCoins(int[] iNums) {
        //prepare to have nums: [1, 3, 1, 5, 8, 1]
        int[] nums = new int[iNums.length + 2];
        int n = 1;
        for (int x : iNums) if (x > 0) nums[n++] = x;
        nums[0] = nums[n++] = 1;

        //memo[i][j]: memorize the max coin value when burst from left to right
        int[][] memo = new int[n][n];
        return burst(memo, nums, 0, n - 1);
    }

    //return max coin from bursting ballon[left] to ballon[right] (exclude left and right)
    public int burst(int[][] memo, int[] nums, int left, int right) {
        if (left + 1 == right) return 0;
        //if already memorize, no need to calculate, return
        if (memo[left][right] > 0) return memo[left][right];
        int ans = 0;
        for (int i = left + 1; i < right; ++i)
            ans = Math.max(ans, 
                    nums[left] * nums[i] * nums[right]  //burst balloon i last
                  + burst(memo, nums, left, i) //max coin from bursting ballon[left] to ballon[i]
                  + burst(memo, nums, i, right)); //max coin from bursting ballon[i] to ballon[right]
        memo[left][right] = ans;
        return ans;
    }
}

Reference

Leetcode