LeetCode - Stack - Remove Outermost Parentheses 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 "" + "" = "". public String removeOuterParentheses(String S) { String result = ""; Stack<String> s1 = new Stack<String>(); Stack<String> s2 = new Stack<String>(); for(int i=0;i<S.length();i++){ //System.out.println(S.charAt(i)); if(s1.size() == 0){ s1.push(S.charAt(i)+""); } else if(s1.size() > 0){ if(("(").equals(S.charAt(i)+"")){ s2.push(S.charAt(i)+""); result = result+S.charAt(i); } else if(s2.size() > 0 && (")").equals(S.charAt(i)+"")){ s2.pop(); result = result+S.charAt(i); } else{ s1.pop(); } } } return result; }