312. Burst Balloons

📋 Đề Bài

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 i^th 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

🧠 Thuật Toán & Kỹ Thuật

Dynamic Programming (Quy hoạch động)

⏱️ Thời gian O(n)

💾 Không gian O(n)

💻 Lời Giải

Python 0312-burst-balloons.py

class Solution:
    def maxCoins(self, nums: List[int]) -> int:
        nums = [1] + nums + [1]
        n = len(nums)
        memo = {}
        
        def dp(i, j):
            if i >= j:
                return 0
            if (i, j) in memo:
                return memo[(i, j)]
            ans = 0
            for k in range(i + 1, j):
                ans = max(ans, nums[i] * nums[k] * nums[j] + dp(i, k) + dp(k, j))
            memo[(i, j)] = ans
            return ans
        
        return dp(0, n - 1)