<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:
    string makeLargestSpecial(string s) {
        int i = 0;
        return dfs(s, i);
    }

    string dfs(string& s, int& i) {
        string res;
        vector<string> toks;
        while (i < s.size() && res.empty()) {
            if (s[i++] == '1') toks.push_back(dfs(s, i));
            else res += "1";
        }
        bool prefix = res.size();
        sort(toks.begin(), toks.end());
        for (int i = toks.size() - 1; i >= 0; i--) {
            res += toks[i];
        }
        if (prefix) res += '0';
        return res;
    }
};
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