LeetCode #1190 — MEDIUM

Reverse Substrings Between Each Pair of Parentheses

Move from brute-force thinking to an efficient approach using stack strategy.

Solve on LeetCode

The Problem

Problem Statement

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Stack

Example 1

"(abcd)"

Example 2

"(u(love)i)"

Example 3

"(ed(et(oc))el)"

Step 02

Core Insight

What unlocks the optimal approach

Find all brackets in the string.
Does the order of the reverse matter ?
The order does not matter.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1190: Reverse Substrings Between Each Pair of Parentheses
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();
    }
}

// Accepted solution for LeetCode #1190: Reverse Substrings Between Each Pair of Parentheses
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)
}

# Accepted solution for LeetCode #1190: Reverse Substrings Between Each Pair of Parentheses
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)

// Accepted solution for LeetCode #1190: Reverse Substrings Between Each Pair of Parentheses
struct Solution;

use std::collections::VecDeque;

impl Solution {
    fn reverse_parentheses(s: String) -> String {
        let mut stack: Vec<char> = vec![];
        let mut queue: VecDeque<char> = VecDeque::new();
        for c in s.chars() {
            if c == ')' {
                while let Some(top) = stack.pop() {
                    if top == '(' {
                        while let Some(front) = queue.pop_front() {
                            stack.push(front);
                        }
                        break;
                    } else {
                        queue.push_back(top);
                    }
                }
            } else {
                stack.push(c);
            }
        }
        stack.iter().collect()
    }
}

#[test]
fn test() {
    let s = "(abcd)".to_string();
    let res = "dcba".to_string();
    assert_eq!(Solution::reverse_parentheses(s), res);
    let s = "(u(love)i)".to_string();
    let res = "iloveu".to_string();
    assert_eq!(Solution::reverse_parentheses(s), res);
    let s = "(ed(et(oc))el)".to_string();
    let res = "leetcode".to_string();
    assert_eq!(Solution::reverse_parentheses(s), res);
    let s = "a(bcdefghijkl(mno)p)q".to_string();
    let res = "apmnolkjihgfedcbq".to_string();
    assert_eq!(Solution::reverse_parentheses(s), res);
}

// Accepted solution for LeetCode #1190: Reverse Substrings Between Each Pair of Parentheses
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('');
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n^2)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan left (or right) to find the next greater/smaller element. The inner scan can visit up to n elements per outer iteration, giving O(n²) total comparisons. No extra space needed beyond loop variables.

MONOTONIC STACK

O(n) time

O(n) space

Each element is pushed onto the stack at most once and popped at most once, giving 2n total operations = O(n). The stack itself holds at most n elements in the worst case. The key insight: amortized O(1) per element despite the inner while-loop.

Shortcut: Each element pushed once + popped once → O(n) amortized. The inner while-loop does not make it O(n²).

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Breaking monotonic invariant

Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.

Usually fails on: Indices point to blocked elements and outputs shift.

Fix: Pop while invariant is violated before pushing current element.