<p><strong>Special binary strings</strong> are binary strings with the following two properties:</p>

<ul>
	<li>The number of <code>0</code>&#39;s is equal to the number of <code>1</code>&#39;s.</li>
	<li>Every prefix of the binary string has at least as many <code>1</code>&#39;s as <code>0</code>&#39;s.</li>
</ul>

<p>You are given a <strong>special binary</strong> string <code>s</code>.</p>

<p>A move consists of choosing two consecutive, non-empty, special substrings of <code>s</code>, 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.</p>

<p>Return <em>the lexicographically largest resulting string possible after applying the mentioned operations on the string</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;11011000&quot;
<strong>Output:</strong> &quot;11100100&quot;
<strong>Explanation:</strong> The strings &quot;10&quot; [occuring at s[1]] and &quot;1100&quot; [at s[3]] are swapped.
This is the lexicographically largest string possible after some number of swaps.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;10&quot;
<strong>Output:</strong> &quot;10&quot;
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 50</code></li>
	<li><code>s[i]</code> is either <code>&#39;0&#39;</code> or <code>&#39;1&#39;</code>.</li>
	<li><code>s</code> is a special binary string.</li>
</ul>

class Solution:
    def makeLargestSpecial(self, s: str) -> str:
        # Base case: if the string is empty, return it
        if not s:
            return ""
        
        substrings = []
        balance = 0
        start = 0

        # Split the string into special substrings
        for i in range(len(s)):
            if s[i] == '1':
                balance += 1
            else:
                balance -= 1
            
            # When balance is zero, we've found a valid special substring
            if balance == 0:
                # Recursively sort the inner substring
                substrings.append('1' + self.makeLargestSpecial(s[start+1:i]) + '0')
                start = i + 1

        # Sort all found special substrings in lexicographically descending order
        substrings.sort(reverse=True)
        
        # Join them into the largest possible string and return
        return ''.join(substrings)
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