Introduction

In this blog post, we will explore how to solve the LeetCode problem "761" 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

Special binary strings are binary strings with the following two properties: The number of 0's is equal to the number of 1's. Every prefix of the binary string has at least as many 1's as 0's. You are given a special binary string s. A move consists of choosing two consecutive, non-empty, special substrings of s, and swapping them. Two strings are consecutive if the last character of the first string is exactly one index before the first character of the second string. Return the lexicographically largest resulting string possible after applying the mentioned operations on the string. Example 1: Input: s = "11011000" Output: "11100100" Explanation: The strings "10" [occuring at s[1]] and "1100" [at s[3]] are swapped. This is the lexicographically largest string possible after some number of swaps. Example 2: Input: s = "10" Output: "10" Constraints: 1 <= s.length <= 50 s[i] is either '0' or '1'. s is a special binary string.

Explanation

Here's a breakdown of the solution:

• Decomposition into Special Substrings: The core idea is to recursively decompose the given special string into its constituent special substrings. This involves identifying balanced '1' and '0' sections that maintain the special string property.
• Lexicographical Ordering: Once we have a list of special substrings, we sort them in reverse lexicographical order. This ensures that the "largest" special substrings are placed at the beginning of the rearranged string.

• Recursive Rearrangement: The sorting is applied recursively within each special substring, ensuring optimal arrangement at all levels.

• Complexity: Time Complexity: O(n^2) in the worst case due to recursion and sorting. Space Complexity: O(n) due to recursion depth and storing substrings.

Code

    class Solution:
    def makeLargestSpecial(self, s: str) -> str:
        """
        Decomposes the special string into special substrings, sorts them in reverse lexicographical order,
        and recursively applies the process within each substring.
        """

        if not s:
            return ""

        count = 0
        start = 0
        substrings = []

        for i in range(len(s)):
            if s[i] == '1':
                count += 1
            else:
                count -= 1

            if count == 0:
                substrings.append('1' + self.makeLargestSpecial(s[start + 1:i]) + '0')
                start = i + 1

        substrings.sort(reverse=True)
        return "".join(substrings)