Burst balloons

class Solution:
    def maxCoins(self, nums: List[int]) -> int:
        n = len(nums)
        points = [0] * (n + 2) # left and right dummy balloons
        points[0], points[n + 1] = 1, 1
        for i in range(1, n + 1):
            points[i] = nums[i - 1]

        dp = [[0] * (n + 2) for _ in range(n + 2)]

        for i in range(n, -1, -1):
            for j in range(i + 1, n + 2):
                for k in range(i + 1, j):
                    dp[i][j] = max(
                        dp[i][j],
                        dp[i][k] + dp[k][j] + points[i] * points[j] * points[k]
                    )

        return dp[0][n + 1]

Burst Balloons

Difficulty: Hard

You are given n balloons, indexed from 0 to n - 1. Each balloon is painted with a number on it represented by an array nums. You are asked to burst all the balloons.

If you burst the ith balloon, you will get nums[i - 1] * nums[i] * nums[i + 1] coins. If i - 1 or i + 1 goes out of bounds of the array, then treat it as if there is a balloon with a 1 painted on it.

Return the maximum coins you can collect by bursting the balloons wisely.

 

Example 1:

Input: nums = [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

Example 2:

Input: nums = [1,5]
Output: 10

 

Constraints:

  • n == nums.length
  • 1 <= n <= 300
  • 0 <= nums[i] <= 100