<p>Given a string&nbsp;<code>s</code><var>,</var>&nbsp;return <em>the maximum&nbsp;number of unique substrings that the given string can be split into</em>.</p>

<p>You can split string&nbsp;<code>s</code> into any list of&nbsp;<strong>non-empty substrings</strong>, where the concatenation of the substrings forms the original string.&nbsp;However, you must split the substrings such that all of them are <strong>unique</strong>.</p>

<p>A <strong>substring</strong> is a contiguous sequence of characters within a string.</p>

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

<pre>
<strong>Input:</strong> s = &quot;ababccc&quot;
<strong>Output:</strong> 5
<strong>Explanation</strong>: One way to split maximally is [&#39;a&#39;, &#39;b&#39;, &#39;ab&#39;, &#39;c&#39;, &#39;cc&#39;]. Splitting like [&#39;a&#39;, &#39;b&#39;, &#39;a&#39;, &#39;b&#39;, &#39;c&#39;, &#39;cc&#39;] is not valid as you have &#39;a&#39; and &#39;b&#39; multiple times.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;aba&quot;
<strong>Output:</strong> 2
<strong>Explanation</strong>: One way to split maximally is [&#39;a&#39;, &#39;ba&#39;].
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;aa&quot;
<strong>Output:</strong> 1
<strong>Explanation</strong>: It is impossible to split the string any further.
</pre>

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

<ul>
	<li>
	<p><code>1 &lt;= s.length&nbsp;&lt;= 16</code></p>
	</li>
	<li>
	<p><code>s</code> contains&nbsp;only lower case English letters.</p>
	</li>
</ul>

class Solution {
public:
    int maxUniqueSplit(string s) {
        int ret = 1, n = s.size(), i, mask;
        for(int fences=0; fences < (1 << (n - 1)); fences++) {
            if(__builtin_popcount(fences) < ret)
                continue;
            unordered_set<string> st;
            string tmp = {s[0]};
            for(i=0; i<n-1; i++) {
                if((1<<i) & fences) {
                    st.insert(tmp);
                    tmp = {s[i + 1]};
                }
                else
                    tmp.push_back(s[i + 1]);
            }
            if(tmp.size() > 0)
                st.insert(tmp);
            ret = max(ret, (int)st.size());
        }
        return ret;
    }  
};
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