Codility - Lesson 7 Stacks and Queues - 3. Nesting

Source Link:
https://codility.com/programmers/lessons/7-stacks_and_queues/nesting/

Question:

A string S consisting of N characters is called properly nested if:

• S is empty;
• S has the form "(U)" where U is a properly nested string;
• S has the form "VW" where V and W are properly nested strings.

For example, string "(()(())())" is properly nested but string "())" isn't.
Write a function:

class Solution { public int solution(String S); }

that, given a string S consisting of N characters, returns 1 if string S is properly nested and 0 otherwise.

For example, given S = "(()(())())", the function should return 1 and given S = "())", the function should return 0, as explained above.

Assume that:

• N is an integer within the range [0..1,000,000];
• string S consists only of the characters "(" and/or ")".

Complexity:

• expected worst-case time complexity is O(N);
• expected worst-case space complexity is O(1) (not counting the storage required for input arguments).

My Solution:
Notes:
1. special case 1: empty string
// special case 2: odd length
else if( S.length() % 2 == 1)
return 0; 
2. main idea: use "stack" to check
Stack<Character> st = new Stack<>();
3.
if( S.charAt(i)=='(' ){
st.push(')'); // note: push its pair (be careful)
} 
4.
else if(S.charAt(i)==')'){
// before pop: need to check if stack is empty (important)
if(st.isEmpty() == true){
return 0;
}
else{
char temp = st.pop();
if( temp != ')'){
return 0;
}
}
}
} 
5. be careful: if stack is "not empty" -> return 0
if( !st.isEmpty() )
return 0;
else  
return 1;