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Special Binary String

Difficulty: Hard

Problem Description

You are given a special binary string s. A move consists of choosing two consecutive, non-empty, special substrings of s, and swapping them. The goal is to return the lexicographically largest resulting string possible after applying the mentioned operations on the string.

Key Insights

  • Special binary strings have equal numbers of 0s and 1s.
  • Every prefix of a special binary string has at least as many 1s as 0s.
  • The problem involves identifying special substrings and determining the best way to swap them to achieve the largest lexicographical order.
  • A recursive approach can be employed to break down the string into special substrings, sort them, and then combine them back together.

Space and Time Complexity

Time Complexity: O(n log n) - due to sorting of substrings. Space Complexity: O(n) - for storing the substrings and final result.

Solution

The solution involves using a recursive approach to break down the special binary string into its special substrings. We identify and extract these substrings, sort them in reverse lexicographical order, and then concatenate them to form the final result. This guarantees that the largest possible string is formed by utilizing the properties of special binary strings.

Code Solutions

def makeLargestSpecial(s: str) -> str:
    count = i = 0
    special_substrings = []
    
    # Identify special substrings
    for j in range(len(s)):
        count += 1 if s[j] == '1' else -1
        if count == 0:  # Found a special substring
            # Recursively process the inner substring
            special_substrings.append('1' + makeLargestSpecial(s[i + 1:j]) + '0')
            i = j + 1

    # Sort the special substrings in reverse lexicographical order
    special_substrings.sort(reverse=True)
    
    # Concatenate and return the result
    return ''.join(special_substrings)

# Example usage
print(makeLargestSpecial("11011000"))  # Output: "11100100"
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