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;
    }

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