Reduce string by removing outermost parentheses from each primitive substring
Last Updated :
03 May, 2023
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 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 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# 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[c] == '(' &&
count++ > 0)
// Include the character
res += S[c];
// If closing parentheses is
// encountered and their
// count is less than count
// of opening parentheses
if (S[c] == ')' &&
count-- > 1)
// Include the character
res += S[c];
}
// 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
<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>
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;
}
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
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# 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
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.
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