Reverse substrings between each pair of parenthesis
Given a string str that consists of lower case 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: geeksforgeeksInput: 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 stringstring 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 codeint main(){ string str = "(skeeg(for)skeeg)"; int len = str.length(); cout << reverseParentheses(str, len); return 0;} |
Java
// Java implementation of the approachimport 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 stringdef 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 substarting # after the last encountered opening # bracket till the current character elif (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 codeif __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 approachusing 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 stringstatic 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 codestatic 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 approachfunction 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 stringfunction 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 codelet str = "(skeeg(for)skeeg)";let len = str.length;document.write(reverseParentheses(str, len));// This code is contributed by patel2127</script> |
Output:
geeksforgeeks
