956. Tallest Billboard

You are installing a billboard and want it to have the largest height. The billboard will have two steel supports, one on each side. Each steel support must be an equal height.

You have a collection of rods which can be welded together. For example, if you have rods of lengths 1, 2, and 3, you can weld them together to make a support of length 6.

Return the largest possible height of your billboard installation. If you cannot support the billboard, return 0.

Example 1:

Input: [1,2,3,6]
Output: 6
Explanation: We have two disjoint subsets {1,2,3} and {6}, which have the same sum = 6.

Example 2:

Input: [1,2,3,4,5,6]
Output: 10
Explanation: We have two disjoint subsets {2,3,5} and {4,6}, which have the same sum = 10.

Example 3:

Input: [1,2]
Output: 0
Explanation: The billboard cannot be supported, so we return 0.

Note:

0 <= rods.length <= 20
1 <= rods[i] <= 1000
The sum of rods is at most 5000.

思路：暴力BFS TLE；改用dp，dp[i][j]=True，表示（I,J）2部分可以分别得到i，j，具体求解过程就是对于没来一个数，就更新一遍dp[i][j]数组，但是复杂度太大了。

仔细想想，每来rods数组中一个数，是用来改变（I,J）之间差值，所以其实只需要一维数组就够了，dp[i]表示（I,J）差值为i时，（I,J）中小的/大的那个数最大是多少

But don't forget, we don't care every state. We only care about the best state.
For example, if we have dp[i][j] = True and dp[i+k][j+k] = True. Do we still care dp[i][j] ? No!

So the state description can be simplized to dp[dx] = len, which means we can realize the state (len, len+dx).

class Solution(object):
    def tallestBillboard(self, rods):
        """
        :type rods: List[int]
        :rtype: int
        """
        if not rods: return 0
        su = sum(rods)
        dp = [-1]*(su+1)
        dp[0] = 0
        dp2=list(dp)
        
        for i in rods:
            for j in range(su+1):
                if dp[j]==-1: continue
                
                # add to larger
                if j+i<=su:
                    dp2[j+i]=max(dp2[j+i], dp[j])
                
                # add to smaller
                if j>=i:
                    dp2[j-i]=max(dp2[j-i], dp[j]+i)
                else:
                    dp2[i-j]=max(dp2[i-j], dp[j]+j)
            
            dp=list(dp2)
                        
        return dp[0]
        
    
s=Solution()
print(s.tallestBillboard([61,45,43,54,40,53,55,47,51,59,42]))
print(s.tallestBillboard([2,4,8,16]))
print(s.tallestBillboard([1,2,3,6]))
#print(s.tallestBillboard([1,2,3,4,5,6]))
#print(s.tallestBillboard([1,2]))

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