[LeetCode]#1021. Remove Outermost Parentheses

Environment: Python 3.7

Key technique: enumerate

A valid parentheses string is either empty (""), "(" + A + ")", or A + B, where A and B are valid parentheses strings, and + represents string concatenation. For example, "", "()", "(())()", and "(()(()))" are all valid parentheses strings.

A valid parentheses string S is primitive if it is nonempty, and there does not exist a way to split it into S = A+B, with A and B nonempty valid parentheses strings.

Given a valid parentheses string S, consider its primitive decomposition: S = P_1 + P_2 + ... + P_k, where P_i are primitive valid parentheses strings.

Return S after removing the outermost parentheses of every primitive string in the primitive decomposition of S.

Example 1:

Input: "(()())(())"
Output: "()()()"
Explanation: 
The input string is "(()())(())", with primitive decomposition "(()())" + "(())".
After removing outer parentheses of each part, this is "()()" + "()" = "()()()".

Example 2:

Input: "(()())(())(()(()))"
Output: "()()()()(())"
Explanation: 
The input string is "(()())(())(()(()))", with primitive decomposition "(()())" + "(())" + "(()(()))".
After removing outer parentheses of each part, this is "()()" + "()" + "()(())" = "()()()()(())".

Example 3:

Input: "()()"
Output: ""
Explanation: 
The input string is "()()", with primitive decomposition "()" + "()".
After removing outer parentheses of each part, this is "" + "" = "".

Note:

1. S.length <= 10000
2. S[i] is "(" or ")"
3. S is a valid parentheses string

Analysis:

1. As below, record second “(“ as start point. If count is 0, output location start+1 to 0 previous location.
2. Add 1 to current location (count=0) as next start location.

Solution:

class Solution:
    def removeOuterParentheses(self, S):
        ans = ""
        start = 0
        zero =0       
        for i, s in enumerate(S):
            if s == "(":
                zero += 1
            else:
                zero -= 1
            if  zero == 0: 
                ans += S[start+1:i]
                start = i+1
        return ans

Submitted result:

Reference:

https://medium.com/@gulaer0812/leetcode-%E8%A9%B3%E8%A7%A3-1021-remove-outermost-parentheses-python3-e938ea7d0223