1190. Reverse Substrings Between Each Pair of Parentheses

Problem Description

You are given a string s that consists of lower case English letters and brackets.

Reverse the strings in each pair of matching parentheses, starting from the innermost one.

Your result should not contain any brackets.

Examples

Example 1:

Input: s = "(u(love)i)"
Output: "iloveu"
Explanation: The substring "love" is reversed first, then the whole string is reversed.

Constraints

• 1 <= s.length <= 2000
• s only contains lower case English characters and parentheses.
• It is guaranteed that all parentheses are balanced.

Solution for 1190. Reverse Substrings Between Each Pair of Parentheses

Solution

Implementation

Live Editor

function Solution(arr) {
    var reverseParentheses = function(s) {
    let st = [];
    let arr = s.split('');
    
    for (let i = 0; i < arr.length; i++) {
        if (arr[i] === '(') {
            st.push(i);
        } else if (arr[i] === ')') {
            let top = st.pop();
            reverse(arr, top + 1, i - 1);
        }
    }
    
    let ans = '';
    for (let c of arr) {
        if (/[a-z]/.test(c)) {
            ans += c;
        }
    }
    return ans;
};

function reverse(arr, left, right) {
    while (left < right) {
        let temp = arr[left];
        arr[left] = arr[right];
        arr[right] = temp;
        left++;
        right--;
    }
}

  const input = "(u(love)i)"
  const output = reverseParentheses(input)
  return (
    <div>
      <p>
        <b>Input: </b>
        {JSON.stringify(input)}
      </p>
      <p>
        <b>Output:</b> {output.toString()}
      </p>
    </div>
  );
}

Result

Loading...

Complexity Analysis

• Time Complexity: $O(n^2)$
• Space Complexity: $O(n)$

Code in Different Languages

Python

Written by @hiteshgahanolia

class Solution:
 def reverseParentheses(self, s: str) -> str:
     st = []
     s = list(s)
     
     for i in range(len(s)):
         if s[i] == '(':
             st.append(i)
         elif s[i] == ')':
             top = st.pop()
             s[top + 1:i] = s[top + 1:i][::-1]
     
     return ''.join(c for c in s if c.isalpha())

Java

Written by @hiteshgahanolia

import java.util.Stack;

public class Solution {
 public String reverseParentheses(String s) {
     Stack<Integer> st = new Stack<>();
     char[] charArray = s.toCharArray();
     
     for (int i = 0; i < charArray.length; i++) {
         if (charArray[i] == '(') {
             st.push(i);
         } else if (charArray[i] == ')') {
             int top = st.pop();
             reverse(charArray, top + 1, i - 1);
         }
     }
     
     StringBuilder ans = new StringBuilder();
     for (char c : charArray) {
         if (Character.isLetter(c)) {
             ans.append(c);
         }
     }
     return ans.toString();
 }
 
 private void reverse(char[] array, int left, int right) {
     while (left < right) {
         char temp = array[left];
         array[left] = array[right];
         array[right] = temp;
         left++;
         right--;
     }
 }
}

C++

Written by @hiteshgahanolia

class Solution {
public:
 string reverseParentheses(string s) {
     stack<int>st;
     for(int i=0;i<s.size();i++)
     {
         if(s[i]=='(')
         {
             st.push(i);
         }
         else if(s[i]==')')
         {
             int top = st.top();
             st.pop();
             reverse(s.begin()+top , s.begin()+i);
         }
     }

     string ans="";
     for(int i=0;i<s.size();i++)
     {
         if(s[i]>='a' && s[i]<='z')
         {
             ans+=s[i];
         }
     }
     return ans;
 }
};

References

• LeetCode Problem: 1190. Reverse Substrings Between Each Pair of Parentheses

• Solution Link: LeetCode Solution