Reverse substrings between each pair of parenthesis

Given a string str that consists of lowercase English letters and brackets. The task is to reverse the substrings in each pair of matching parentheses, starting from the innermost one. The result should not contain any brackets.

Examples:

Input: str = “(skeeg(for)skeeg)” 
Output: geeksforgeeks

Input: str = “((ng)ipm(ca))” 
Output: camping 

Approach: This problem can be solved using a stack. First, whenever a ‘(‘ is encountered then push the index of the element into the stack, and whenever a ‘)’ is encountered then get the top element of the stack as the latest index and reverse the string between the current index and index from the top of the stack. Follow this for the rest of the string and finally print the updated string.

Below is the implementation of the above approach:  

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the modified string
string reverseParentheses(string str, int len)
{
    stack<int> st;

    for (int i = 0; i < len; i++) {

        // Push the index of the current
        // opening bracket
        if (str[i] == '(') {
            st.push(i);
        }

        // Reverse the substring starting
        // after the last encountered opening
        // bracket till the current character
        else if (str[i] == ')') {
            reverse(str.begin() + st.top() + 1,
                    str.begin() + i);
            st.pop();
        }
    }

    // To store the modified string
    string res = "";
    for (int i = 0; i < len; i++) {
        if (str[i] != ')' && str[i] != '(')
            res += (str[i]);
    }
    return res;
}

// Driver code
int main()
{
    string str = "(skeeg(for)skeeg)";
    int len = str.length();

    cout << reverseParentheses(str, len);

    return 0;
}

Java

// Java implementation of the approach
import java.io.*;
import java.util.*;
class GFG {
  static void reverse(char A[], int l, int h)
  {
    if (l < h)
    {
      char ch = A[l];
      A[l] = A[h];
      A[h] = ch;
      reverse(A, l + 1, h - 1);
    }
  }

  // Function to return the modified string
  static String reverseParentheses(String str, int len)
  {
    Stack<Integer> st = new Stack<Integer>();
    for (int i = 0; i < len; i++)
    {

      // Push the index of the current
      // opening bracket
      if (str.charAt(i) == '(')
      {
        st.push(i);
      }

      // Reverse the substring starting
      // after the last encountered opening
      // bracket till the current character
      else if (str.charAt(i) == ')')
      {
        char[] A = str.toCharArray();
        reverse(A, st.peek() + 1, i);
        str = String.copyValueOf(A);
        st.pop();
      }
    }

    // To store the modified string
    String res = "";
    for (int i = 0; i < len; i++)
    {
      if (str.charAt(i) != ')' && str.charAt(i) != '(')
      {
        res += (str.charAt(i));
      }
    }
    return res;
  }

  // Driver code
  public static void main (String[] args)
  {
    String str = "(skeeg(for)skeeg)";
    int len = str.length();
    System.out.println(reverseParentheses(str, len));
  }
}

// This code is contributed by avanitrachhadiya2155

Python3

# Python3 implementation of the approach

# Function to return the modified string
def reverseParentheses(strr, lenn):
    st = []

    for i in range(lenn):

        # Push the index of the current
        # opening bracket
        if (strr[i] == '('):
            st.append(i)

        # Reverse the substring starting
        # after the last encountered opening
        # bracket till the current character
        else if (strr[i] == ')'):
            temp = strr[st[-1]:i + 1]
            strr = strr[:st[-1]] + temp[::-1] + \
                   strr[i + 1:]
            del st[-1]

    # To store the modified string
    res = ""
    for i in range(lenn):
        if (strr[i] != ')' and strr[i] != '('):
            res += (strr[i])
    return res

# Driver code
if __name__ == '__main__':
    strr = "(skeeg(for)skeeg)"
    lenn = len(strr)
    st = [i for i in strr]

    print(reverseParentheses(strr, lenn))

# This code is contributed by Mohit Kumar

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;
using System.Text;

class GFG{

static void reverse(char[] A, int l, int h)
{
    if (l < h)
    {
        char ch = A[l];
        A[l] = A[h];
        A[h] = ch;
        reverse(A, l + 1, h - 1);
    }
}

// Function to return the modified string
static string reverseParentheses(string str, int len)
{
    Stack<int> st = new Stack<int>();

    for(int i = 0; i < len; i++)
    {

        // Push the index of the current
        // opening bracket
        if (str[i] == '(')
        {
            st.Push(i);
        }

        // Reverse the substring starting
        // after the last encountered opening
        // bracket till the current character
        else if (str[i] == ')')
        {
            char[] A = str.ToCharArray();
            reverse(A, st.Peek() + 1, i);
            str = new string(A);
            st.Pop();
        }
    }

    // To store the modified string
    string res = "";
    for(int i = 0; i < len; i++)
    {
        if (str[i] != ')' && str[i] != '(')
        {
            res += str[i];
        }
    }
    return res;
}

// Driver code
static public void Main()
{
    string str = "(skeeg(for)skeeg)";
    int len = str.Length;

    Console.WriteLine(reverseParentheses(str, len));
}
}

// This code is contributed by rag2127

Javascript

<script>
// Javascript implementation of the approach
function reverse(A,l,h)
{
    if (l < h)
    {
      let ch = A[l];
      A[l] = A[h];
      A[h] = ch;
      reverse(A, l + 1, h - 1);
    }
}

 // Function to return the modified string
function reverseParentheses(str,len)
{
    let st = [];
    for (let i = 0; i < len; i++)
    {

      // Push the index of the current
      // opening bracket
      if (str[i] == '(')
      {
        st.push(i);
      }

      // Reverse the substring starting
      // after the last encountered opening
      // bracket till the current character
      else if (str[i] == ')')
      {

        let A = [...str]
        reverse(A, st[st.length-1] + 1, i);
        str = [...A];
        st.pop();
      }
    }

    // To store the modified string
    let res = "";
    for (let i = 0; i < len; i++)
    {
      if (str[i] != ')' && str[i] != '(')
      {
        res += (str[i]);
      }
    }
    return res;
}

 // Driver code
let str = "(skeeg(for)skeeg)";
let len = str.length;
document.write(reverseParentheses(str, len));

// This code is contributed by patel2127
</script>

Output:

geeksforgeeks

Time Complexity: O(n²), where n is the length of the given string.
Auxiliary Space: O(n)