<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:
    void unique(unordered_set<string> us, string s, int p, int c, int *v) {
        int n = s.size();
        if(p == n) {
            if(*v < c)
                *v = c;
            return;
        }
        for(int l = 1; l <= n-p; l++) {
            string sub = s.substr(p, l);
            if (us.find(sub) == us.end()) {
                us.insert(sub);
                unique(us, s, p+l, c+1, v);
                us.erase(sub);
            }
        }
    }

    int maxUniqueSplit(string s) {
        unordered_set<string> us;
        int ans = 0;
        unique(us, s, 0, 0, &ans);
        return ans;
    }
};
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