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.
from functools import lru_cache
class Solution:
def maxCoins(self, nums: List[int]) -> int:
# reframe the problem
nums = [1] + nums + [1]
# cache this
@lru_cache(None)
def dp(left, right):
if left + 1 == right: return 0
# add each balloon on the interval and return the maximum score
return max(nums[left] * nums[i] * nums[right] + dp(left, i) + dp(i, right) for i in range(left+1, right))
# find the maximum number of coins obtained from adding all balloons from (0, len(nums) - 1)
return dp(0, len(nums)-1)