Hard — Dynamic Programming | Array | Interval DP

The Problem

Given an array of balloons with coin values, find the maximum coins you can collect by bursting all balloons, where bursting a balloon gives coins equal to the product of its value and its adjacent neighbors' values.

Approach

Uses dynamic programming with interval DP approach. Adds virtual balloons with value 1 at both ends, then for each subarray, tries all possible last balloons to burst and takes the maximum coins achievable.

Time: O(n³) · Space: O(n²)

Code

class Solution:
 def maxCoins(self, nums: List[int]) -> int:
 n = len(nums)

 # Add virtual balloons at the beginning and end with a value of 1.
 nums = [1] + nums + [1]

 # Create a 2D table dp to store the maximum coins.
 dp = [[0] * (n + 2) for _ in range(n + 2)]

 # Iterate through different balloon ranges.
 for length in range(2, n + 2):
 for left in range(0, n + 2 - length):
 right = left + length
 for k in range(left + 1, right):
 # Choose the best order to burst balloons in the range [left, right].
 dp[left][right] = max(
 dp[left][right],
 nums[left] * nums[k] * nums[right] + dp[left][k] + dp[k][right],
 )

 return dp[0][n + 1]

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