Introduction

In this blog post, we will explore how to solve the LeetCode problem "1021" 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 valid parentheses string is either empty "", "(" + A + ")", or A + B, where A and B are valid parentheses strings, and + represents string concatenation. For example, "", "()", "(())()", and "(()(()))" are all valid parentheses strings. A valid parentheses string s is primitive if it is nonempty, and there does not exist a way to split it into s = A + B, with A and B nonempty valid parentheses strings. Given a valid parentheses string s, consider its primitive decomposition: s = P1 + P2 + ... + Pk, where Pi are primitive valid parentheses strings. Return s after removing the outermost parentheses of every primitive string in the primitive decomposition of s. Example 1: Input: s = "(()())(())" Output: "()()()" Explanation: The input string is "(()())(())", with primitive decomposition "(()())" + "(())". After removing outer parentheses of each part, this is "()()" + "()" = "()()()". Example 2: Input: s = "(()())(())(()(()))" Output: "()()()()(())" Explanation: The input string is "(()())(())(()(()))", with primitive decomposition "(()())" + "(())" + "(()(()))". After removing outer parentheses of each part, this is "()()" + "()" + "()(())" = "()()()()(())". Example 3: Input: s = "()()" Output: "" Explanation: The input string is "()()", with primitive decomposition "()" + "()". After removing outer parentheses of each part, this is "" + "" = "". Constraints: 1 <= s.length <= 105 s[i] is either '(' or ')'. s is a valid parentheses string.

Explanation

Here's the breakdown of the solution:

• Balance Tracking: Use a balance counter to keep track of the nesting level of parentheses. Increment for '(' and decrement for ')'.
• Primitive Decomposition: Identify primitive strings by detecting when the balance returns to zero. The substrings between balance resets are the primitive components.

• Outer Removal: Construct the result by appending the primitive string excluding the outermost parentheses.

• Runtime Complexity: O(n), where n is the length of the input string. Storage Complexity: O(n) due to the stringbuilder.

Code

    def remove_outer_parentheses(s: str) -> str:
    """
    Removes the outermost parentheses of every primitive string in the primitive decomposition of s.
    """
    balance = 0
    start = 0
    result = ""
    for i, char in enumerate(s):
        if char == '(':
            balance += 1
        else:
            balance -= 1

        if balance == 0:
            result += s[start + 1:i]
            start = i + 1

    return result