How do you solve Maximum Nesting Depth of the Parentheses on LeetCode?

Maximum Nesting Depth of the Parentheses (LeetCode #1614) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The maximum depth is determined by counting the number of nested parentheses.

What is the brute force solution for Maximum Nesting Depth of the Parentheses?

The brute force approach for Maximum Nesting Depth of the Parentheses is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves scanning through the string and counting the depth of parentheses at each character. For every opening parenthesis '(', we increase the depth, and for every closing parenthesis ')', we decrease it. We keep track of the maximum depth encountered during this process. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Nesting Depth of the Parentheses?

Maximum Nesting Depth of the Parentheses uses the following concepts: String, Stack. The recommended patterns to study are: Stack, Greedy.

What are the interview tips for Maximum Nesting Depth of the Parentheses?

Always explain your thought process while coding. Consider edge cases, such as strings with no parentheses. Practice dry-running your code with sample inputs to ensure correctness.

What are common mistakes when solving Maximum Nesting Depth of the Parentheses?

Forgetting to update max_depth after increasing current_depth. Incorrectly handling the current_depth when encountering closing parentheses.

#1614

Maximum Nesting Depth of the Parentheses

Easy

StringStackStackGreedy

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

The optimal solution is similar to the brute force approach but focuses on a single pass through the string while maintaining the current depth and maximum depth. This approach ensures we only traverse the string once, making it more efficient.

⚙️

Algorithm

5 steps

1Step 1: Initialize `current_depth` and `max_depth` to 0.
2Step 2: Loop through each character in the string.
3Step 3: Increase `current_depth` for '(', and update `max_depth` if necessary.
4Step 4: Decrease `current_depth` for ')'.
5Step 5: Return `max_depth` at the end.

solution.py10 lines

1def maxDepth(s):
2    current_depth = 0
3    max_depth = 0
4    for char in s:
5        if char == '(':  
6            current_depth += 1
7            max_depth = max(max_depth, current_depth)
8        elif char == ')':
9            current_depth -= 1
10    return max_depth

ℹ

Complexity note: The optimal solution runs in O(n) time since we only make a single pass through the string, and we use O(1) space as we only need a few variables to track the current and maximum depth.

1The maximum depth is determined by counting the number of nested parentheses.
2Each opening parenthesis increases depth, while each closing parenthesis decreases it.

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