[LeetCode] 312. Burst Balloons_hard tag: 区间Dynamic Programming

Given n balloons, indexed from 0 to n-1. Each balloon is painted with a number on it represented by array nums. You are asked to burst all the balloons. If the you burst balloon i you will get nums[left] * nums[i] * nums[right] coins. Here left and right are adjacent indices of i. After the burst, the left and right then becomes adjacent.

Find the maximum coins you can collect by bursting the balloons wisely.

Note:

• You may imagine nums[-1] = nums[n] = 1. They are not real therefore you can not burst them.
• 0 ≤ n ≤ 500, 0 ≤ nums[i] ≤ 100

Example:

Input: [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

 

这个题目的思路也是利用 区间Dynamic Programming, 思想跟[LeetCode] 877. Stone Game == [LintCode] 396. Coins in a Line 3_hard tag: 区间Dynamic Programming, 博弈类似, 

动态表达式为: 

dp[i][j] 是得到的 [i,j] 里面气球的最大值

for k in [i,j]:

　　value_m = ballons[i-1] * ballons[k] * ballons[j+1]

　　dp[i][j] = max(dp[i][j], dp[i][k-1] + value_m + dp[k+1][j])

init:

dp[i][i] = ballons[i-1]* ballons[i]* ballons[i+1]

dp[i][j]  = 0 if j < i

 

1. Constraints

1) size [0, 500]

2) element [0, 100]

 

2. Ideas

 

Dynamic Programming      T: O(n^2)    S; O(n^2)

3. Code

class Solution:
    def burstBallons(self, nums):
        n = len(nums)
        ballons = [1] + nums + [1] # plus the corner case
        dp = [[0]*(n+2) for _ in range(n+2)] # for corner case
        flag = [[0]*(n+2) for _ in range(n+2)]
        def helper(l, r):
            if flag[l][r]:
                return dp[l][r]
            if l == r:
                dp[l][r] = ballons[l-1] * ballons[l] * ballons[l+1]
            elif l < r:
                for k in range(l, r+1): # k belongs [l, r], so it is r + 1
                    value_m = ballons[l-1] * ballons[k] * ballons[r+1]
                    value_l = helper(l, k-1)
                    value_r = helper(k+1, r)
                    dp[l][r] = max(dp[l][r], value_l + value_m + value_r)
            flag[l][r] = 1
            return dp[l][r]
        return helper(1, n)