<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 n = s.length();
        int ans = 1;
        for (int i = 0; i < (1 << (n - 1)); i++) {
            set<string> se;
            vector<int> sep = {-1};
            for (int j = 0; j < n - 1; j++) {
                if ((i >> j) & 1) sep.push_back(j);
            }
            sep.push_back(n - 1);
            bool ok = true;
            for (int i = 1; i < (int)sep.size(); i++) {
                string cur = s.substr(sep[i - 1] + 1, sep[i] - sep[i - 1]);
                if (se.count(cur)) {
                    ok = false;
                    break;
                }
                se.insert(cur);
            }
            if (ok) {
                ans = max(ans, (int)sep.size() - 1);
            }
        }
        return ans;
    }
};
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