##### 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
```