Leetcode 312. Burst Balloons

Problem

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 numsi - 1 * numsi * numsi + 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.

Algorithm

Dynamic Programming (DP). Let dpst represent the maximum number of coins that can be obtained by bursting all the balloons between index s and t.
d p s t = m a x ( d p s k + d p k t + n u m s s ∗ n u m s k ∗ n u m s t ) dpst = max(dpsk + dpkt + numss * numsk * numst) dpst=max(dpsk+dpkt+numss∗numsk∗numst)

For s < k < t s < k < t s<k<t.

Code

class Solution:
    def maxCoins(self, nums: List[int]) -> int:
        nums = [1] + nums + [1]
        nlen = len(nums)
        dp = [[0] * nlen for _ in range(nlen)]
        
        for l in range(2, nlen):
            for s in range(nlen - l):
                t = s + l
                for k in range(s+1, t):
                    if dp[s][t] < dp[s][k] + dp[k][t] + nums[s] * nums[k] * nums[t]:
                        dp[s][t] = dp[s][k] + dp[k][t] + nums[s] * nums[k] * nums[t]
        
        return dp[0][nlen - 1]