Reverse substrings between each pair of parenthesis
Last Updated :
18 Oct, 2023
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++
#include <bits/stdc++.h>
using namespace std;
string reverseParentheses(string str, int len)
{
stack<int> st;
for (int i = 0; i < len; i++) {
if (str[i] == '(') {
st.push(i);
}
else if (str[i] == ')') {
reverse(str.begin() + st.top() + 1,
str.begin() + i);
st.pop();
}
}
string res = "";
for (int i = 0; i < len; i++) {
if (str[i] != ')' && str[i] != '(')
res += (str[i]);
}
return res;
}
int main()
{
string str = "(skeeg(for)skeeg)";
int len = str.length();
cout << reverseParentheses(str, len);
return 0;
}
|
Java
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);
}
}
static String reverseParentheses(String str, int len)
{
Stack<Integer> st = new Stack<Integer>();
for (int i = 0; i < len; i++)
{
if (str.charAt(i) == '(')
{
st.push(i);
}
else if (str.charAt(i) == ')')
{
char[] A = str.toCharArray();
reverse(A, st.peek() + 1, i);
str = String.copyValueOf(A);
st.pop();
}
}
String res = "";
for (int i = 0; i < len; i++)
{
if (str.charAt(i) != ')' && str.charAt(i) != '(')
{
res += (str.charAt(i));
}
}
return res;
}
public static void main (String[] args)
{
String str = "(skeeg(for)skeeg)";
int len = str.length();
System.out.println(reverseParentheses(str, len));
}
}
|
Python3
def reverseParentheses(strr, lenn):
st = []
for i in range(lenn):
if (strr[i] == '('):
st.append(i)
else if (strr[i] == ')'):
temp = strr[st[-1]:i + 1]
strr = strr[:st[-1]] + temp[::-1] + \
strr[i + 1:]
del st[-1]
res = ""
for i in range(lenn):
if (strr[i] != ')' and strr[i] != '('):
res += (strr[i])
return res
if __name__ == '__main__':
strr = "(skeeg(for)skeeg)"
lenn = len(strr)
st = [i for i in strr]
print(reverseParentheses(strr, lenn))
|
C#
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);
}
}
static string reverseParentheses(string str, int len)
{
Stack<int> st = new Stack<int>();
for(int i = 0; i < len; i++)
{
if (str[i] == '(')
{
st.Push(i);
}
else if (str[i] == ')')
{
char[] A = str.ToCharArray();
reverse(A, st.Peek() + 1, i);
str = new string(A);
st.Pop();
}
}
string res = "";
for(int i = 0; i < len; i++)
{
if (str[i] != ')' && str[i] != '(')
{
res += str[i];
}
}
return res;
}
static public void Main()
{
string str = "(skeeg(for)skeeg)";
int len = str.Length;
Console.WriteLine(reverseParentheses(str, len));
}
}
|
Javascript
<script>
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 reverseParentheses(str,len)
{
let st = [];
for (let i = 0; i < len; i++)
{
if (str[i] == '(')
{
st.push(i);
}
else if (str[i] == ')')
{
let A = [...str]
reverse(A, st[st.length-1] + 1, i);
str = [...A];
st.pop();
}
}
let res = "";
for (let i = 0; i < len; i++)
{
if (str[i] != ')' && str[i] != '(')
{
res += (str[i]);
}
}
return res;
}
let str = "(skeeg(for)skeeg)";
let len = str.length;
document.write(reverseParentheses(str, len));
</script>
|
Time Complexity: O(n2), where n is the length of the given string.
Auxiliary Space: O(n)
Efficient Approach:
Our Approach is simple and we can reduce the space complexity from O(n) to O(1) that is constant space .
Approach:
We create a function for reversing substrings between the opening and closing brackets one by one, beginning with the innermost one. We can use a while loop to keep doing this until we run out of brackets.
In each iteration of the while loop, we first find the index of the string’s initial closing bracket. We exit the loop if there is no closing bracket. Otherwise, we search backwards from the closing bracket to identify the appropriate starting bracket.
The reverseSubstring function is then used to reverse the substring between the opening and closing brackets. Finally, we use the erase function to delete the opening and closing brackets from the string.
Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
string reverseSubstring(string str, int start, int end) {
reverse(str.begin()+start, str.begin()+end+1);
return str;
}
string reverseParentheses(string str) {
int n = str.length();
int start, end;
while (true) {
end = str.find(')', 0);
if (end == string::npos) {
break;
}
start = str.rfind('(', end);
str = reverseSubstring(str, start+1, end-1);
str.erase(start, 1);
str.erase(end-1, 1);
}
return str;
}
int main() {
string str="(skeeg(for)skeeg)";
cout<<reverseParentheses(str)<<endl;
return 0;
}
|
Java
import java.util.Stack;
public class GFG {
static String ReverseSubstring(String str, int start,
int end)
{
StringBuilder reversed = new StringBuilder();
for (int i = end; i >= start; i--) {
reversed.append(str.charAt(i));
}
return str.substring(0, start) + reversed.toString()
+ str.substring(end + 1);
}
static String ReverseParentheses(String str)
{
int start, end;
while (true) {
end = str.indexOf(')');
if (end == -1) {
break;
}
start = str.lastIndexOf('(', end);
str = ReverseSubstring(str, start + 1, end - 1);
str = str.substring(0, start)
+ str.substring(start + 1, end)
+ str.substring(end + 1);
}
return str;
}
public static void main(String[] args)
{
String str = "(skeeg(for)skeeg)";
System.out.println(ReverseParentheses(str));
}
}
|
Python3
def reverse_substring(s, start, end):
return s[:start] + s[start:end+1][::-1] + s[end+1:]
def reverse_parentheses(s):
while True:
end = s.find(')')
if end == -1:
break
start = s.rfind('(', 0, end)
s = reverse_substring(s, start+1, end-1)
s = s[:start] + s[start+1:]
s = s[:end-1] + s[end:]
return s
if __name__ == "__main__":
s = "(skeeg(for)skeeg)"
print(reverse_parentheses(s))
|
C#
using System;
using System.Text;
class Program
{
static string ReverseSubstring(string str, int start, int end)
{
StringBuilder reversed = new StringBuilder();
for (int i = end; i >= start; i--)
{
reversed.Append(str[i]);
}
return str.Substring(0, start) + reversed.ToString() + str.Substring(end + 1);
}
static string ReverseParentheses(string str)
{
int start, end;
while (true)
{
end = str.IndexOf(')');
if (end == -1)
{
break;
}
start = str.LastIndexOf('(', end);
str = ReverseSubstring(str, start + 1, end - 1);
str = str.Substring(0, start) + str.Substring(start + 1, end - start - 1) + str.Substring(end + 1);
}
return str;
}
static void Main(string[] args)
{
string str = "(skeeg(for)skeeg)";
Console.WriteLine(ReverseParentheses(str));
}
}
|
Javascript
function reverseSubstring(str, start, end) {
return str.slice(0, start) +
str.slice(start, end+1).split('').reverse().join('') +
str.slice(end+1);
}
function reverseParentheses(str) {
let start, end;
while (true) {
end = str.indexOf(')');
if (end === -1) {
break;
}
start = str.lastIndexOf('(', end);
str = reverseSubstring(str, start+1, end-1);
str = str.slice(0, start) + str.slice(start+1, end) + str.slice(end+1);
}
return str;
}
let str = "(skeeg(for)skeeg)";
console.log(reverseParentheses(str));
|
Time Complexity: O(n^2), where n is the length of the given string.
Auxiliary Space: O(1),No extra space is used.
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