Reduce string by removing outermost parentheses from each primitive substring ❤️

Given a string S of valid parentheses “(“ and “)”, the task is to print the string obtained by removing the outermost parentheses of every primitive substring from S.

A valid parentheses substring S is primitive if it is non-empty, and cannot be split into two or more non-empty substrings which are also a valid parentheses.

Examples:

Input: S = “(()())(())()”
Output: ()()()
Explanation: The input string is “(()())(())()” can be decomposed into primitive substrings “(()())” + “(())”+”()”. After removing outermost parentheses of each primitive substrings, the string obtained is “()()” + “()” = “()()()”

Input: S = “((()())(())(()(())))”
Output: (()())(())(()(()))

Approach: Follow the steps below to solve the problem:

Initialize a variable count to store the number of opening parentheses, i.e. ‘(‘.
Add every ‘(‘ to the result if count is greater than 0, i.e. add all ‘(‘ after the first ‘(‘ of a primitive substring is encountered.
Add every ‘)’ to the result if count is greater than 1, i.e. add all ‘)’ before the last ‘)’ of a primitive substring is encountered.
Finally, print the resultant string obtained.

Below is the implementation of the above approach-

C++

// C++ program to implement the
// above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to remove the outermost
// parentheses of every primitive
// substring from the given string
string removeOuterParentheses(string S)
{
 
    // Stores the resultant string
    string res;
 
    // Stores the count of
    // opened parentheses
    int count = 0;
 
    // Traverse the string
    for (char c : S) {
 
        // If opening parentheses is
        // encountered and their
        // count exceeds 0
        if (c == '(' && count++ > 0)
 
            // Include the character
            res += c;
 
        // If closing parentheses is
        // encountered and their
        // count is less than count
        // of opening parentheses
        if (c == ')' && count-- > 1)
 
            // Include the character
            res += c;
    }
 
    // Return the resultant string
    return res;
}
 
// Driver Code
int main()
{
    string S = "(()())(())()";
    cout << removeOuterParentheses(S);
}

Java

// Java program to implement the
// above approach
import java.io.*;
class GFG{
   
// Function to remove the outermost
// parentheses of every primitive
// substring from the given string
static String removeOuterParentheses(String S)
{
  // Stores the resultant 
  // string
  String res = "";
 
  // Stores the count of
  // opened parentheses
  int count = 0;
 
  // Traverse the string
  for (int c = 0; 
           c < S.length(); c++) 
  {
    // If opening parentheses is
    // encountered and their
    // count exceeds 0
    if (S.charAt(c) == '(' && 
        count++ > 0)
 
      // Include the character
      res += S.charAt(c);
 
    // If closing parentheses is
    // encountered and their
    // count is less than count
    // of opening parentheses
    if (S.charAt(c) == ')' && 
        count-- > 1)
 
      // Include the character
      res += S.charAt(c);
  }
 
  // Return the resultant string
  return res;
}
 
// Driver Code
public static void main(String[] args)
{
  String S = "(()())(())()";
  System.out.print(removeOuterParentheses(S));
}
}
 
// This code is contributed by Chitranayal

Python3

# Python3 program to implement the
# above approach
 
# Function to remove the outermost
# parentheses of every primitive
# substring from the given string
def removeOuterParentheses(S):
     
    # Stores the resultant string
    res = ""
 
    # Stores the count of
    # opened parentheses
    count = 0
 
    # Traverse the string
    for c in S:
         
        # If opening parentheses is
        # encountered and their
        # count exceeds 0
        if (c == '(' and count > 0):
 
            # Include the character
            res += c
 
        # If closing parentheses is
        # encountered and their
        # count is less than count
        # of opening parentheses
        if (c == '('):
            count += 1
        if (c == ')' and count > 1):
 
            # Include the character
            res += c
             
        if (c == ')'):
          count -= 1
     
    # Return the resultant string
    return res
 
# Driver Code
if __name__ == '__main__':
     
    S = "(()())(())()"
     
    print(removeOuterParentheses(S))
 
# This code is contributed by SURENDRA_GANGWAR

C#

// C# program to implement
// the above approach  
using System;
 
class GFG{
    
// Function to remove the outermost
// parentheses of every primitive
// substring from the given string
static string removeOuterParentheses(string S)
{
   
  // Stores the resultant 
  // string
  string res = "";
  
  // Stores the count of
  // opened parentheses
  int count = 0;
  
  // Traverse the string
  for(int c = 0; c < S.Length; c++) 
  {
     
    // If opening parentheses is
    // encountered and their
    // count exceeds 0
    if (S == '(' && 
        count++ > 0)
  
      // Include the character
      res += S;
  
    // If closing parentheses is
    // encountered and their
    // count is less than count
    // of opening parentheses
    if (S == ')' && 
        count-- > 1)
  
      // Include the character
      res += S;
  }
  
  // Return the resultant string
  return res;
}
  
// Driver Code
public static void Main()
{
  string S = "(()())(())()";
   
  Console.Write(removeOuterParentheses(S));
}
}
 
// This code is contributed by sanjoy_62

Javascript

<script>
 
// Javascript program to implement the
// above approach
 
// Function to remove the outermost
// parentheses of every primitive
// substring from the given string
function removeOuterParentheses(S)
{
 
// Stores the resultant
// string
let res = "";
 
// Stores the count of
// opened parentheses
let count = 0;
 
// Traverse the string
for (let c = 0;
        c < S.length; c++)
{
    // If opening parentheses is
    // encountered and their
    // count exceeds 0
    if (S.charAt(c) == '(' &&
        count++ > 0)
 
    // Include the character
    res += S.charAt(c);
 
    // If closing parentheses is
    // encountered and their
    // count is less than count
    // of opening parentheses
    if (S.charAt(c) == ')' &&
        count-- > 1)
 
    // Include the character
    res += S.charAt(c);
}
 
// Return the resultant string
return res;
}
 
// Driver Code
 
let S = "(()())(())()";
document.write(removeOuterParentheses(S));
 
// This code is contributed by jana_sayantan.
</script>

Output

()()()

Time Complexity: O(N) where n is number of elements in given string. As, we are using a loop to traverse N times so it will cost us O(N) time
Auxiliary Space: O(N), as we are using extra space for string.

The Optimal approach to remove the outermost parentheses from a string can be achieved using a simple algorithm that keeps track of the number of opening and closing parentheses encountered. Here is an optimal approach:

Initialize two variables, open_count and close_count, to zero
Initialize an empty string called result.
Loop through each character c in the input string s.
If c is an opening parenthesis, increment open_count.
If c is a closing parenthesis, increment close_count.
If open_count and close_count are equal and greater than zero, this means that we have encountered a complete pair of opening and closing parentheses, so we can add the substring between them to the result string.
Reset open_count and close_count to zero.
Return the result string.

Here is an implementation of this algorithm:

C++

// C++ Program for the above approach
#include <iostream>
#include <string>
using namespace std;
 
// function to remove outer parentheses
string removeOuterParentheses(string s) {
    int openCount = 0;
    int closeCount = 0;
    string result = "";
    int start = 0;
 
    for (int i = 0; i < s.length(); i++) {
        char c = s[i];
        if (c == '(') {
            openCount++;
        } else if (c == ')') {
            closeCount++;
        }
 
        if (openCount == closeCount) {
            result += s.substr(start+1, i-start-1);
            start = i+1;
        }
    }
    // return the output string(result)
    return result;
}
 
// driver program to test above function
int main() {
    // Example 1
    string s1 = "(()())(())()";
    cout << removeOuterParentheses(s1) << endl;  
 
    // Example 2
    string s2 = "()()(()())(()())";
    cout << removeOuterParentheses(s2) << endl;  
 
    // Example 3
    string s3 = "((()))(())";
    cout << removeOuterParentheses(s3) << endl;  
 
    return 0;
}

Java

public class Main {
    public static String removeOuterParentheses(String s) {
        int openCount = 0;
        int closeCount = 0;
        String result = "";
        int start = 0;
 
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if (c == '(') {
                openCount++;
            } else if (c == ')') {
                closeCount++;
            }
 
            if (openCount == closeCount) {
                result += s.substring(start+1, i);
                start = i+1;
            }
        }
 
        return result;
    }
 
    public static void main(String[] args) {
        // Example 1
        String s1 = "(()())(())()";
        System.out.println(removeOuterParentheses(s1));  
 
        // Example 2
        String s2 = "()()(()())(()())";
        System.out.println(removeOuterParentheses(s2));  
 
        // Example 3
        String s3 = "((()))(())";
        System.out.println(removeOuterParentheses(s3));  
    }
}
 
// This code is contributed by Sundaram

Python3

def remove_outer_parentheses(s):
    open_count = 0
    close_count = 0
    result = ""
    start = 0
 
    for i, c in enumerate(s):
        if c == "(":
            open_count += 1
        elif c == ")":
            close_count += 1
 
        if open_count == close_count:
            result += s[start+1:i]
            start = i+1
 
    return result
 
# Driver code
if __name__ == "__main__":
    # Example 1
    s1 = "(()())(())()"
    print(remove_outer_parentheses(s1))  
 
    # Example 2
    s2 = "()()(()())(()())"
    print(remove_outer_parentheses(s2))  
 
    # Example 3
    s3 = "((()))(())"
    print(remove_outer_parentheses(s3))

C#

// C# Program for the above approach
using System;
 
namespace RemoveOuterParentheses {
class Program {
    // function to remove outer parentheses
    static string RemoveOuterParentheses(string s)
    {
        int openCount = 0;
        int closeCount = 0;
        string result = "";
        int start = 0;
        for (int i = 0; i < s.Length; i++) {
            char c = s[i];
            if (c == '(') {
                openCount++;
            }
            else if (c == ')') {
                closeCount++;
            }
 
            if (openCount == closeCount) {
                result += s.Substring(start + 1,
                                      i - start - 1);
                start = i + 1;
            }
        }
        // return the output string(result)
        return result;
    }
 
    // driver program to test above function
    static void Main(string[] args)
    {
        // Example 1
        string s1 = "(()())(())()";
        Console.WriteLine(RemoveOuterParentheses(s1));
 
        // Example 2
        string s2 = "()()(()())(()())";
        Console.WriteLine(RemoveOuterParentheses(s2));
 
        // Example 3
        string s3 = "((()))(())";
        Console.WriteLine(RemoveOuterParentheses(s3));
    }
}
}
// This code is contributed by sarojmcy2w

Javascript

function remove_outer_parentheses(s) {
    let open_count = 0;
    let close_count = 0;
    let result = "";
    let start = 0;
 
    for (let i = 0; i < s.length; i++) {
        let c = s[i];
        if (c == "(") {
            open_count += 1;
        } else if (c == ")") {
            close_count += 1;
        }
 
        if (open_count == close_count) {
            result += s.slice(start + 1, i);
            start = i + 1;
        }
    }
 
    return result;
}
 
// Driver code
// Example 1
let s1 = "(()())(())()";
console.log(remove_outer_parentheses(s1));
 
// Example 2
let s2 = "()()(()())(()())";
console.log(remove_outer_parentheses(s2));
 
// Example 3
let s3 = "((()))(())";
console.log(remove_outer_parentheses(s3));

Output

()()()
()()()()
(())()

This approach has a time complexity of O(n), where n is the length of the input string s. It uses constant space, except for the output string, which is proportional to the number of valid parentheses pairs in s.

Reduce string by removing outermost parentheses from each primitive substring

C++

Java

Python3

C#

Javascript

C++

Java

Python3

C#

Javascript

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