How do you solve Remove Outermost Parentheses on LeetCode?

Remove Outermost Parentheses (LeetCode #1021) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding the structure of valid parentheses strings is crucial.

What is the brute force solution for Remove Outermost Parentheses?

The brute force approach for Remove Outermost Parentheses is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves iterating through the string to identify and remove the outermost parentheses of each primitive substring. This is straightforward but inefficient as it requires multiple passes over the string. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Remove Outermost Parentheses?

Remove Outermost Parentheses uses the following concepts: String, Stack. The recommended patterns to study are: Stack, Two Pointers.

What are the interview tips for Remove Outermost Parentheses?

Explain your thought process clearly as you code. Consider edge cases, such as empty strings or strings with no outer parentheses. Practice similar problems to become familiar with patterns in parentheses problems.

What are common mistakes when solving Remove Outermost Parentheses?

Not keeping track of the balance correctly. Attempting to manipulate the string in a way that leads to incorrect results.

#1021

Remove Outermost Parentheses

Easy

StringStackStackTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal approach involves a single pass through the string while maintaining a balance counter. This allows us to directly construct the result without needing to track each primitive substring separately.

⚙️

Algorithm

5 steps

1Step 1: Initialize an empty result string and a balance counter.
2Step 2: Iterate through the string, adjusting the balance for each '(' and ')'.
3Step 3: Skip the first '(' and the last ')' of each primitive substring by checking the balance.
4Step 4: Append valid characters to the result.
5Step 5: Return the result string.

solution.py13 lines

1def removeOuterParentheses(s):
2    result = ''
3    balance = 0
4    for char in s:
5        if char == '(':  
6            if balance > 0:
7                result += char
8            balance += 1
9        else:
10            balance -= 1
11            if balance > 0:
12                result += char
13    return result

ℹ

Complexity note: This complexity is linear because we traverse the string once, and the space complexity is linear due to the result string being built.

1Understanding the structure of valid parentheses strings is crucial.
2Recognizing the balance of parentheses helps in identifying primitive substrings.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.