Introduction

In this blog post, we will explore how to solve the LeetCode problem "1111" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

A string is a valid parentheses string (denoted VPS) if and only if it consists of "(" and ")" characters only, and: It is the empty string, or It can be written as AB (A concatenated with B), where A and B are VPS's, or It can be written as (A), where A is a VPS. We can similarly define the nesting depth depth(S) of any VPS S as follows: depth("") = 0 depth(A + B) = max(depth(A), depth(B)), where A and B are VPS's depth("(" + A + ")") = 1 + depth(A), where A is a VPS. For example, "", "()()", and "()(()())" are VPS's (with nesting depths 0, 1, and 2), and ")(" and "(()" are not VPS's. Given a VPS seq, split it into two disjoint subsequences A and B, such that A and B are VPS's (and A.length + B.length = seq.length). Now choose any such A and B such that max(depth(A), depth(B)) is the minimum possible value. Return an answer array (of length seq.length) that encodes such a choice of A and B: answer[i] = 0 if seq[i] is part of A, else answer[i] = 1. Note that even though multiple answers may exist, you may return any of them. Example 1: Input: seq = "(()())" Output: [0,1,1,1,1,0] Example 2: Input: seq = "()(())()" Output: [0,0,0,1,1,0,1,1] Constraints: 1 <= seq.size <= 10000

Explanation

Here's a breakdown of the solution:

• Alternating Assignment: The core idea is to distribute the parentheses between A and B in a way that balances their nesting depths. A simple way to achieve this is to alternate the assignment of opening parentheses to A and B, and similarly for closing parentheses. This helps prevent either A or B from becoming excessively deep.
• Depth Tracking: To implement the alternating assignment, we use a counter to track the current nesting depth of each subsequence. We assign an opening parenthesis to the subsequence with the smaller current depth and increment its depth. The same logic applies to closing parentheses, but we decrement the depth before assignment.

• Minimizing Max Depth: By alternating the assignments, we inherently try to keep the depths of A and B as close as possible. This strategy leads to a minimization of the maximum depth between the two subsequences.

• Runtime Complexity: O(n) & Storage Complexity: O(n), where n is the length of the input string seq.

Code

    def max_nesting_depth_after_split(seq: str) -> list[int]:
    """
    Splits a valid parentheses string into two disjoint subsequences A and B,
    such that A and B are valid parentheses strings, and
    max(depth(A), depth(B)) is minimized.

    Args:
        seq: A valid parentheses string.

    Returns:
        An answer array (of length seq.length) that encodes the choice of A and B:
        answer[i] = 0 if seq[i] is part of A, else answer[i] = 1.
    """

    a_depth = 0
    b_depth = 0
    result = []

    for char in seq:
        if char == '(':
            if a_depth < b_depth:
                result.append(0)
                a_depth += 1
            else:
                result.append(1)
                b_depth += 1
        else:  # char == ')'
            if a_depth > b_depth:
                result.append(0)
                a_depth -= 1
            else:
                result.append(1)
                b_depth -= 1

    return result