What is the brute force solution for Reverse Substrings Between Each Pair of Parentheses?

The brute force approach for Reverse Substrings Between Each Pair of Parentheses is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves repeatedly searching for the innermost parentheses, reversing the substring within them, and replacing the parentheses with the reversed string. This continues until no parentheses are left in the string. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Reverse Substrings Between Each Pair of Parentheses?

Reverse Substrings Between Each Pair of Parentheses uses the following concepts: String, Stack. The recommended patterns to study are: Stack, Recursion.

What are the interview tips for Reverse Substrings Between Each Pair of Parentheses?

Think about using a stack for problems involving nested structures. Always dry-run your code with sample inputs to catch edge cases.

What are common mistakes when solving Reverse Substrings Between Each Pair of Parentheses?

Not handling nested parentheses correctly. Forgetting to reverse the substring before pushing it back to the stack.

#1190

Reverse Substrings Between Each Pair of Parentheses

Medium

StringStackStackRecursion

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

Using a stack allows us to efficiently manage the nested structure of parentheses. We push characters onto the stack until we find a closing parenthesis, at which point we pop from the stack until we reach the corresponding opening parenthesis, reversing the substring in the process.

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Algorithm

5 steps

1Step 1: Initialize an empty stack.
2Step 2: Iterate through each character in the string.
3Step 3: If the character is '(', push it onto the stack.
4Step 4: If the character is ')', pop from the stack until '(' is found, reverse the popped characters, and push the reversed substring back onto the stack.
5Step 5: After processing all characters, join all elements in the stack to form the final result.

solution.py13 lines

1def reverseParentheses(s):
2    stack = []
3    for char in s:
4        if char == ')':
5            temp = ''
6            while stack and stack[-1] != '(': 
7                temp += stack.pop()
8            stack.pop()  # pop the '('
9            for c in temp:
10                stack.append(c)
11        else:
12            stack.append(char)
13    return ''.join(stack)

ℹ

Complexity note: The optimal solution runs in linear time because we process each character once, and the stack operations (push/pop) are constant time. The space complexity is linear due to the stack storing characters.

1Using a stack simplifies handling nested structures.
2Reversing substrings can be efficiently managed with stack operations.

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