burst-balloons

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Description

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 <= 500
• 0 <= nums[i] <= 100

Solution(Python)

class Solution(object):
    def maxCoins(self, nums):
        nums = [1] + nums + [1]
        dp = [[0] * len(nums) for _ in nums]
        n = len(nums)
        for i in range(n - 3, -1, -1):
            for j in range(i + 2, n):
                dp[i][j] = max(
                    [
                        dp[i][k] + dp[k][j] + nums[i] * nums[k] * nums[j]
                        for k in range(i + 1, j)
                    ]
                )
        return dp[0][n - 1]