Short Problem Definition:
Determine whether given string of parentheses is properly nested.
expected worst-case time complexity is
expected worst-case space complexity is
Because there is only one type of brackets, the problem is easier than Brackets. Just check if there is always a opening bracket before a closing one.
def solution(S): leftBrackets = 0 for symbol in S: if symbol == '(': leftBrackets += 1 else: if leftBrackets == 0: return 0 leftBrackets -= 1 if leftBrackets != 0: return 0 return 1