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.

public class Solution {
    public int maxCoins(int[] nums) {
        if (nums == null || nums.length == 0) return 0;

        int[] ballons = new int[nums.length + 2];
        int len = 1;
        for (int num : nums) {
            if (num > 0) {
                ballons[len++] = num;
            }
        }
        ballons[0] = ballons[len++] = 1;
        int[][] coins = new int[len][len];
        for (int k = 3; k <= len; k++) {//there are at least three balloons
            for (int left = 0; left + k - 1 < len; left ++) {
                int right = left + k - 1;
                for (int i = left + 1; i < right; i ++) {
                    coins[left][right] = Math.max(coins[left][right], ballons[left] * ballons[i] * ballons[right] + coins[left][i] + coins[i][right]);
                }
            }
        }
        return coins[0][len - 1];
    }
}

class Solution:
    def maxCoins(self, nums: List[int]) -> int:
        if not nums:
            return 0

        nums = [num for num in nums if num > 0]
        nums = [1] + nums + [1]
        dp = [[0] * len(nums) for _ in range(len(nums))]
        for i in range(3, len(nums) + 1):
            for left in range(len(nums) - i + 1):
                right = left + i - 1
                for j in range(left + 1, right):
                    dp[left][right] = max(dp[left][right], nums[left] * nums[j] * nums[right] + dp[left][j] + dp[j][right])
        return dp[0][-1]