What is the key dynamic programming pattern in Burst Balloons?

The problem uses state transition DP where each balloon is considered as the last to burst in a subarray, allowing optimal coin calculation.

How do I handle edge balloons when calculating coins?

Treat out-of-bound neighbors as 1 by padding the array at both ends, simplifying calculations for first and last balloons.

Is recursion sufficient or should I use iterative DP?

Memoized recursion works, but iterative DP ensures all subarrays are processed systematically and avoids stack overflows for large n.

Why is time complexity O(n^3) for Burst Balloons?

Each of the O(n^2) subarrays requires iterating O(n) possible last-burst positions, giving total O(n^3) operations.

Can this DP approach handle arrays up to length 300 efficiently?

Yes, with proper memoization or DP table filling, O(n^3) operations are feasible for n up to 300 without exceeding time limits.

#312

Hard

auto_awesome状态·转移·动态规划

LeetCode 题解工作台

戳气球

有 n 个气球，编号为 0 到 n - 1 ，每个气球上都标有一个数字，这些数字存在数组 nums 中。现在要求你戳破所有的气球。戳破第 i 个气球，你可以获得 nums[i - 1] * nums[i] * nums[i + 1] 枚硬币。这里的 i - 1 和 i + 1 代表和 i 相邻的…

数组动态规划

题目描述

有 n 个气球，编号为0 到 n - 1，每个气球上都标有一个数字，这些数字存在数组 nums 中。

现在要求你戳破所有的气球。戳破第 i 个气球，你可以获得 nums[i - 1] * nums[i] * nums[i + 1] 枚硬币。这里的 i - 1 和 i + 1 代表和 i 相邻的两个气球的序号。如果 i - 1或 i + 1 超出了数组的边界，那么就当它是一个数字为 1 的气球。

求所能获得硬币的最大数量。

示例 1：

输入：nums = [3,1,5,8]
输出：167
解释：
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

示例 2：

输入：nums = [1,5]
输出：10

提示：

n == nums.length
1 <= n <= 300
0 <= nums[i] <= 100

lightbulb

解题思路

方法一：动态规划

我们记数组 $nums$ 的长度为 $n$ 。根据题目描述，我们可以在数组 $nums$ 的左右两端各添加一个 $1$ ，记为 $arr$ 。

然后，我们定义 $f[i][j]$ 表示戳破区间 $[i, j]$ 内的所有气球能得到的最多硬币数，那么答案即为 $f[0][n+1]$ 。

对于 $f[i][j]$ ，我们枚举区间 $[i, j]$ 内的所有位置 $k$ ，假设 $k$ 是最后一个戳破的气球，那么我们可以得到如下状态转移方程：

f[i][j] = \max(f[i][j], f[i][k] + f[k][j] + arr[i] \times arr[k] \times arr[j])

在实现上，由于 $f[i][j]$ 的状态转移方程中涉及到 $f[i][k]$ 和 $f[k][j]$ ，其中 $i < k < j$ ，因此我们需要从大到小地遍历 $i$ ，从小到大地遍历 $j$ ，这样才能保证当计算 $f[i][j]$ 时 $f[i][k]$ 和 $f[k][j]$ 已经被计算出来。

最后，我们返回 $f[0][n+1]$ 即可。

时间复杂度 $O(n^3)$ ，空间复杂度 $O(n^2)$ 。其中 $n$ 为数组 $nums$ 的长度。

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class Solution:
    def maxCoins(self, nums: List[int]) -> int:
        n = len(nums)
        arr = [1] + nums + [1]
        f = [[0] * (n + 2) for _ in range(n + 2)]
        for i in range(n - 1, -1, -1):
            for j in range(i + 2, n + 2):
                for k in range(i + 1, j):
                    f[i][j] = max(f[i][j], f[i][k] + f[k][j] + arr[i] * arr[k] * arr[j])
        return f[0][-1]

speed

复杂度分析

指标	值
时间	complexity is O(n^3) due to iterating all subarrays and possible last balloons within each. Space complexity is O(n^2) for the DP table storing maximum coins for all ranges.
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Expect discussion of recursive state definition versus iterative DP filling.
question_mark
Look for correct handling of boundary conditions using virtual balloons of value 1.
question_mark
Check awareness of overlapping subproblems and use of memoization or DP table.

warning

常见陷阱

外企场景

error
Forgetting to pad nums with 1 at both ends leading to index errors.
error
Incorrectly defining DP state, such as including endpoints instead of exclusive ranges.
error
Neglecting to iterate all possible last-burst positions within a subarray.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Maximizing coins when balloons can have negative numbers, requiring careful multiplication handling.
arrow_right_alt
Bursting balloons with additional constraints, e.g., only adjacent balloons can be burst next.
arrow_right_alt
Adapting the DP approach to count number of distinct maximum-coin sequences.

help

常见问题

外企场景

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