Short Problem Definition:

Determine whether given string of parentheses is properly nested.

Link

Nesting

Complexity:

expected worst-case time complexity is O(N)

expected worst-case space complexity is O(1)

Execution:

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.

Solution:

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