<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 {
public:
    // Function to make the largest special string
    string makeLargestSpecial(string S) {
        vector<string> specials;
        int count = 0, start = 0;

        // Step 1: Split the string into smaller valid special binary substrings
        for (int i = 0; i < S.size(); ++i) {
            count += (S[i] == '1') ? 1 : -1;

            // When we have an equal number of '1's and '0's, we have found a valid special string
            if (count == 0) {
                // Recursively sort the internal part (S[start+1 : i-1]) and wrap with "1" and "0"
                specials.push_back("1" + makeLargestSpecial(S.substr(start + 1, i - start - 1)) + "0");
                start = i + 1;
            }
        }

        // Step 2: Sort all special substrings in descending order
        sort(specials.begin(), specials.end(), greater<string>());

        // Step 3: Join all the sorted special substrings and return the result
        string result;
        for (const string& special : specials) {
            result += special;
        }

        return result;
    }
};
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