/*
 * @lc app=leetcode id=1021 lang=rust
 *
 * [1021] Remove Outermost Parentheses
 *
 * https://leetcode.com/problems/remove-outermost-parentheses/description/
 *
 * algorithms
 * Easy (76.58%)
 * Total Accepted:    69.1K
 * Total Submissions: 90.2K
 * Testcase Example:  '"(()())(())"'
 *
 * A valid parentheses string is either empty (""), "(" + A + ")", or A + B,
 * where A and B are valid parentheses strings, and + represents string
 * concatenation.  For example, "", "()", "(())()", and "(()(()))" are all
 * valid parentheses strings.
 * 
 * A valid parentheses string S is primitive if it is nonempty, and there does
 * not exist a way to split it into S = A+B, with A and B nonempty valid
 * parentheses strings.
 * 
 * Given a valid parentheses string S, consider its primitive decomposition: S
 * = P_1 + P_2 + ... + P_k, where P_i are primitive valid parentheses strings.
 * 
 * Return S after removing the outermost parentheses of every primitive string
 * in the primitive decomposition of S.
 * 
 * 
 * 
 * Example 1:
 * 
 * 
 * Input: "(()())(())"
 * Output: "()()()"
 * Explanation: 
 * The input string is "(()())(())", with primitive decomposition "(()())" +
 * "(())".
 * After removing outer parentheses of each part, this is "()()" + "()" =
 * "()()()".
 * 
 * 
 * 
 * Example 2:
 * 
 * 
 * Input: "(()())(())(()(()))"
 * Output: "()()()()(())"
 * Explanation: 
 * The input string is "(()())(())(()(()))", with primitive decomposition
 * "(()())" + "(())" + "(()(()))".
 * After removing outer parentheses of each part, this is "()()" + "()" +
 * "()(())" = "()()()()(())".
 * 
 * 
 * 
 * Example 3:
 * 
 * 
 * Input: "()()"
 * Output: ""
 * Explanation: 
 * The input string is "()()", with primitive decomposition "()" + "()".
 * After removing outer parentheses of each part, this is "" + "" = "".
 * 
 * 
 * 
 * 
 * 
 * 
 * Note:
 * 
 * 
 * S.length <= 10000
 * S[i] is "(" or ")"
 * S is a valid parentheses string
 * 
 * 
 * 
 * 
 * 
 * 
 * 
 */
impl Solution {
    pub fn remove_outer_parentheses(s: String) -> String {
        let mut res:String = "".into(); 
        let mut cter = 0; 
        for c in s.chars(){
            if c == ')' { cter -= 1;}
            if cter > 0 { res.push(c);}
            if c == '(' { cter += 1; }
        }
        res 
    }
}
// pub structSolution;