Skip to content

1190. Reverse Substrings Between Each Pair of Parentheses

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.

 

Example 1:

Input: s = "(abcd)"
Output: "dcba"

Example 2:

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

Example 3:

Input: s = "(ed(et(oc))el)"
Output: "leetcode"
Explanation: First, we reverse the substring "oc", then "etco", and finally, the whole string.

 

Constraints:

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

Solutions

Solution 1: Simulation

We can directly use a stack to simulate the reversal process.

The time complexity is $O(n^2)$, and the space complexity is $O(n)$, where $n$ is the length of the string $s$.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
class Solution:
    def reverseParentheses(self, s: str) -> str:
        stk = []
        for c in s:
            if c == ")":
                t = []
                while stk[-1] != "(":
                    t.append(stk.pop())
                stk.pop()
                stk.extend(t)
            else:
                stk.append(c)
        return "".join(stk)

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
class Solution {
    public String reverseParentheses(String s) {
        StringBuilder stk = new StringBuilder();
        for (char c : s.toCharArray()) {
            if (c == ')') {
                StringBuilder t = new StringBuilder();
                while (stk.charAt(stk.length() - 1) != '(') {
                    t.append(stk.charAt(stk.length() - 1));
                    stk.deleteCharAt(stk.length() - 1);
                }
                stk.deleteCharAt(stk.length() - 1);
                stk.append(t);
            } else {
                stk.append(c);
            }
        }
        return stk.toString();
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
class Solution {
public:
    string reverseParentheses(string s) {
        string stk;
        for (char& c : s) {
            if (c == ')') {
                string t;
                while (stk.back() != '(') {
                    t.push_back(stk.back());
                    stk.pop_back();
                }
                stk.pop_back();
                stk += t;
            } else {
                stk.push_back(c);
            }
        }
        return stk;
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
func reverseParentheses(s string) string {
    stk := []byte{}
    for i := range s {
        if s[i] == ')' {
            t := []byte{}
            for stk[len(stk)-1] != '(' {
                t = append(t, stk[len(stk)-1])
                stk = stk[:len(stk)-1]
            }
            stk = stk[:len(stk)-1]
            stk = append(stk, t...)
        } else {
            stk = append(stk, s[i])
        }
    }
    return string(stk)
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
function reverseParentheses(s: string): string {
    const stk: string[] = [];
    for (const c of s) {
        if (c === ')') {
            const t: string[] = [];
            while (stk.at(-1)! !== '(') {
                t.push(stk.pop()!);
            }
            stk.pop();
            stk.push(...t);
        } else {
            stk.push(c);
        }
    }
    return stk.join('');
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
/**
 * @param {string} s
 * @return {string}
 */
var reverseParentheses = function (s) {
    const