<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>

#pragma GCC optimize("Ofast")
static auto _ = [] () {ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);return 0;}();
class Solution {
public:
    int maxUniqueSplit(string s) {
        vector<string> split;
        string tempStr = string(1, s[0]);
        backTracking(s, split, tempStr, 1 , s.size());
        return ans;
    }
private:    
    int ans = 0;
    void backTracking(string s, vector<string>& split, string& curStr, int cur, int len){        
        if (cur > len){            
            ans = max(ans, (int)split.size());
            return;
        }
        curStr.push_back(s[cur]);
        backTracking(s, split, curStr, cur+1, len);
        curStr.pop_back();
        
        if ( find(split.begin(), split.end() ,curStr) == split.end() ){
            string tempStr = string(1, s[cur]);
            split.push_back(curStr);
            backTracking(s, split, tempStr, cur+1, len);
            split.pop_back();
        }
    }    
};
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