761. Special Binary String - Detailed Explanation

1. Problem Statement

A special binary string is a binary string that satisfies the following properties:

• It is non-empty.
• It is balanced: it contains the same number of '1's and '0's.
• For every prefix of the string, the number of '1's is greater than or equal to the number of '0's.

Problem:
Given a special binary string S, you are allowed to perform the following operation any number of times:

• Choose two consecutive, non-empty, special binary substrings in S and swap them.

Return the lexicographically largest string you can obtain after performing these operations.

Example 1

• Input:

  S = "11011000"
  

• Output:

  "11100100"
  

• Explanation:
  The string "11011000" can be partitioned and reassembled by swapping its special substrings. One optimal transformation yields "11100100", which is lexicographically larger than the original.

Example 2

• Input:

  S = "10"
  

• Output:

  "10"
  

• Explanation:
  With only two characters, the string "10" is already the only possible special binary string.

Constraints

• The length of S is between 1 and 50.
• S is guaranteed to be a special binary string.

2. Hints and Key Observations

1. Structure Resembling Valid Parentheses:
   Notice that a special binary string follows the same structure as a valid parentheses string if you consider '1' as '(' and '0' as ')'. This means every special binary string can be uniquely decomposed into “primitive” (or “indecomposable”) special substrings.

2. Decomposition Insight:
   As you scan S, maintain a counter: increment it for '1' and decrement for '0'. Every time the counter returns to 0, you have found a special substring.
   For example, in "11011000", scanning from left to right:

   • Start: count = 0
   • Read '1' → count = 1
   • Read '1' → count = 2
   • Read '0' → count = 1
   • Read '1' → count = 2
   • Read '1' → count = 3
   • Read '0' → count = 2
   • Read '0' → count = 1
   • Read '0' → count = 0
     Here, the entire string is a valid special substring, but it may contain nested special substrings.

3. Lexicographical Order via Recursion:
   The key observation is that if you can recursively process the inner parts of a special substring, then by sorting the resulting pieces in descending order and concatenating them, you achieve the lexicographically largest string.

Detailed Approach

Recursive Decomposition

1. Partition the String:

   • Initialize a counter and an index pointer i = 0.
   • Iterate through the string with an index j.
   • Update the counter: add 1 for '1' and subtract 1 for '0'.
   • When the counter becomes 0, it means the substring S[i:j+1] is a balanced special binary string.

2. Process the Inner Part:

   • Remove the outer characters (the first '1' and the last '0') from the substring.
   • Recursively process the inner substring (S[i+1:j]) to get its lexicographically largest form.

3. Wrap and Collect:

   • Form the string as "1" + <result of recursion> + "0".
   • Collect all such processed substrings.

4. Sort and Concatenate:

   • Sort the collected substrings in descending order.
   • Concatenate them to obtain the final result.

Visual Walkthrough (Example: S = "11011000")

• Step 1:
  Begin scanning "11011000".
  The entire string is a special substring since the counter reaches 0 at the end.

• Step 2:
  Remove the outer '1' and '0' to get the inner substring: "101100".

• Step 3:
  Recursively process "101100" by similarly partitioning it into its special substrings, processing each recursively, and then sorting them in descending order.

• Step 4:
  Finally, wrap each processed substring with '1' and '0', sort them, and join them together to get "11100100".

4. Complexity Analysis

• Time Complexity:
  In the worst case, the algorithm runs in O(n²) due to the recursive decomposition and the sorting at each level (with n ≤ 50, this is acceptable).

• Space Complexity:
  O(n) for recursion and temporary storage.

Code Implementations

Python Implementation