#1111

Medium

auto_awesomeStack-based state management

LeetCode Problem Workspace

Maximum Nesting Depth of Two Valid Parentheses Strings

This problem challenges you to compute the maximum nesting depth of two valid parentheses strings using stack-based state management.

string stack

Problem Statement

Given a string containing valid parentheses, your task is to compute the nesting depth for each position in the string. A valid parentheses string (VPS) consists of '(' and ')', where the nesting depth is the number of opening parentheses surrounding a given character. For example, the string '(()())' has a depth of 1 for each '(' and ')' that is enclosed by parentheses.

Your goal is to calculate the nesting depth at each position in the input string. For each character, output the current depth of the string at that position. The result will be an array that represents the depth of each character in the string. Use a stack-based approach to manage the depth as you iterate over the string.

Examples

Example 1

Input: seq = "(()())"

Output: [0,1,1,1,1,0]

Example details omitted.

Example 2

Input: seq = "()(())()"

Output: [0,0,0,1,1,0,1,1]

Example details omitted.

Constraints

1 <= seq.size <= 10000

Solution Approach

Using Stack for Depth Tracking

We can track the depth of parentheses by using a stack. As we traverse through the string, we push to the stack when encountering an opening parenthesis '(' and pop from the stack when encountering a closing parenthesis ')'. The size of the stack will give the current depth at each point in the string.

Handling Nested Parentheses

When parentheses are nested, the depth increases as you encounter more '(' characters. Each time you encounter a ')', the stack size decreases, reflecting the exit from a nested level. This process is repeated for each character in the string.

Efficient Time and Space Complexity

Given that you only need to traverse the string once and maintain a stack, both the time and space complexity are O(n), where n is the length of the string. This allows for efficient computation even with larger inputs.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of the solution is O(n), where n is the length of the string, since each character is processed once. The space complexity is O(n) as well, since the stack can hold up to n characters in the worst case.

What Interviewers Usually Probe

Check if the candidate understands stack-based state management.
Evaluate how well the candidate optimizes space and time complexity.
Ensure the candidate handles edge cases like consecutive closing parentheses or unbalanced parentheses correctly.

Common Pitfalls or Variants

Common pitfalls

Forgetting to reset the depth value after encountering a closing parenthesis.
Incorrectly handling unbalanced parentheses strings.
Not accounting for deeply nested parentheses and overestimating depth at certain points.

Follow-up variants

Implementing a solution without a stack, using a different data structure or approach.
Allowing for invalid parentheses strings and handling errors or exceptions.
Modifying the problem to return a boolean indicating if the parentheses are valid, instead of tracking depth.

FAQ

What is the primary pattern used to solve the Maximum Nesting Depth of Two Valid Parentheses Strings?

The primary pattern is stack-based state management, where the depth of parentheses is tracked using a stack.

What is the time complexity of the stack-based approach?

The time complexity is O(n), where n is the length of the string, as each character is processed once.

How can I handle invalid parentheses strings in this problem?

This problem assumes the input is always a valid parentheses string. Handling invalid strings would involve additional validation before processing the string.

What is the best data structure to track the nesting depth in this problem?

The stack is the best data structure for this problem as it allows for efficient tracking of the depth as you encounter each parenthesis.

How does GhostInterview help with solving this problem?

GhostInterview provides structured practice, detailed feedback, and hints to help you solve stack-based problems like this one effectively.

terminal

Solution

Solution 1: Greedy

We use a variable $x$ to maintain the current balance of parentheses, which is the number of left parentheses minus the number of right parentheses.

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class Solution:
    def maxDepthAfterSplit(self, seq: str) -> List[int]:
        ans = [0] * len(seq)
        x = 0
        for i, c in enumerate(seq):
            if c == "(":
                ans[i] = x & 1
                x += 1
            else:
                x -= 1
                ans[i] = x & 1
        return ans

Continue Practicing

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