Formatted question description: https://leetcode.ca/all/1190.html

# 1190. Reverse Substrings Between Each Pair of Parentheses (Medium)

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 bracket.

Example 1:

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


Example 2:

Input: s = "(u(love)i)"
Output: "iloveu"


Example 3:

Input: s = "(ed(et(oc))el)"
Output: "leetcode"


Example 4:

Input: s = "a(bcdefghijkl(mno)p)q"
Output: "apmnolkjihgfedcbq"


Constraints:

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

Companies:

Related Topics:
Stack

## Solution 1. Recursive

We scan through the string. Once we see (, we go one layer deeper. Each layer returns once it see a ) or end of string.

// OJ: https://leetcode.com/problems/reverse-substrings-between-each-pair-of-parentheses/

// Time: O(N^2)
// Space: O(N^2)
class Solution {
private:
string solve(string &s, int &i) {
string ans;
int N = s.size();
while (i < N) {
while (i < N && s[i] != '(' && s[i] != ')') ans += s[i++];
if (i < N) ++i;
if (i >= N || s[i - 1] == ')') break;
auto mid = solve(s, i);
reverse(mid.begin(), mid.end());
ans += mid;
}
return ans;
}
public:
string reverseParentheses(string s) {
int i = 0;
return solve(s, i);
}
};


## Solution 2.

There is a pattern in the reversed string. For each pair of parenthesis, when we see '(' or ')', we jump to the corresponding ')' or '(' and toggle the direction we scan.

Example:

   1   2    3   4
v   v    v   v
a  ( b ( cd ) e ) f

• We scan rightwards initially. ans = a
• When we meet the parenthesis 1 from left to right, we jump to its corresponding parenthesis 4 and toggle the direction to be leftwards. ans = ae
• When we meet the parenthesis 3 from right to left, we jump to its corresponding parenthesis 2 and toggle direction to be rightwards. ans = aecd
• When we meet the parenthesis 3 from left to right, we jump to 2 and toggle direction to be leftwards. ans = aecdb.
• When we meet the parenthesis 1 from right to left, we jump to 4 and toggle direction to be rightwards. ans = aecdbf.
• We’ve reached the end. The answer is aecdbf.
// OJ: https://leetcode.com/problems/reverse-substrings-between-each-pair-of-parentheses/

// Time: O(N)
// Space: O(P) where P is the number of pairs of parenthesis in s.
// Ref: https://leetcode.com/problems/reverse-substrings-between-each-pair-of-parentheses/discuss/383670/JavaC%2B%2BPython-Why-not-O(N)
class Solution {
public:
string reverseParentheses(string s) {
stack<int> st;
unordered_map<int, int> m;
for (int i = 0; i < s.size(); ++i) {
if (s[i] == '(') st.push(i);
else if (s[i] == ')') {
m[i] = st.top();
m[st.top()] = i;
st.pop();
}
}
string ans;
for (int i = 0, d = 1; i < s.size(); i += d) {
if (s[i] == '(' || s[i] == ')') {
i = m[i];
d = -d;
} else ans += s[i];
}
return ans;
}
};


Java

class Solution {
public String reverseParentheses(String s) {
Stack<Character> stack = new Stack<Character>();
int length = s.length();
for (int i = 0; i < length; i++) {
char c = s.charAt(i);
if (c != ')')
stack.push(c);
else {
while (!stack.isEmpty() && stack.peek() != '(')
queue.offer(stack.pop());
stack.pop();
while (!queue.isEmpty())
stack.push(queue.poll());
}
}
StringBuffer sb = new StringBuffer();
while (!stack.isEmpty())
sb.append(stack.pop());
sb.reverse();
return sb.toString();
}
}