
import java.util.Stack


/**
 * 100/100
 * @param S
 * @return
 */
fun solution(S: String): Int {
    if (S.length == 0) return 1
    if (S.length % 2 != 0) return 0
    val stack: Stack<Char> = Stack<Char>()
    for (i in 0 until S.length) {
        val character = S[i]
        if (character == ')') {
            if (!stack.isEmpty() && stack.peek() == '(') {
                stack.pop()
            } else {
                return 0
            }
        } else {
            stack.push(character)
        }
    }
    return if (stack.size == 0) {
        1
    } else 0
}

/**
 * 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.
 *
 * Write an efficient algorithm for the following assumptions:
 *
 * N is an integer within the range [0..1,000,000];
 * string S consists only of the characters "(" and/or ")".
 */