動機

題目寫得很爛，但題目本身很有趣!!

題目想做的事是把string的括號分成兩組，讓兩組的深度越小越好 回傳每個括號是分配到哪一組

Problem

A string is a valid parentheses string (denoted VPS) if and only if it consists of "(" and ")" characters only, and:

  • It is the empty string, or
  • It can be written as AB (A concatenated with B), where A and B are VPS's, or
  • It can be written as (A), where A is a VPS.

We can similarly define the nesting depth depth(S) of any VPS S as follows:

  • depth("") = 0
  • depth(A + B) = max(depth(A), depth(B)), where A and B are VPS's
  • depth("(" + A + ")") = 1 + depth(A), where A is a VPS.

For example,  "", "()()", and "()(()())" are VPS's (with nesting depths 0, 1, and 2), and ")(" and "(()" are not VPS's.

 

Given a VPS seq, split it into two disjoint subsequences A and B, such that A and B are VPS's (and A.length + B.length = seq.length).

Now choose any such A and B such that max(depth(A), depth(B)) is the minimum possible value.

Return an answer array (of length seq.length) that encodes such a choice of A and B:  answer[i] = 0 if seq[i] is part of A, else answer[i] = 1.  Note that even though multiple answers may exist, you may return any of them.

 

Example 1:

Input: seq = "(()())"Output: [0,1,1,1,1,0]

Example 2:

Input: seq = "()(())()"Output: [0,0,0,1,1,0,1,1]

 

Constraints:

  • 1 <= seq.size <= 10000

Sol

就是把括號分成兩半，依據深度!!

class Solution:
    def maxDepthAfterSplit(self, seq: str) -> List[int]:
        ret = []
        d = 0
        for c in seq:
            left = (c == "(")
            
            if left:
                d += 1
            
            ret += int(d % 2 == 0),
            
            if not left:
                d -= 1
        return ret