<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 print(unordered_map<string,int>&mp){
    for(auto it=mp.begin();it!=mp.end();it++){
        cout<<it->first<<' ';
    }
    cout<<endl;
}
int solve(int i, string &str, unordered_map<string,int>&mp,int j){
    if(i>=str.length()){
        // print(mp);
         return 0;
         }
    if(j>=str.length()) return -1e9;
    // break
    int b=0;
    if( mp.find(str.substr(i,j-i+1))==mp.end())
    {
        mp[str.substr(i,j-i+1)]++;
        b=max(b, 1+ solve(j+1,str, mp, j+1));
        mp.erase(str.substr(i,j-i+1));
    }

    // not break

    return max(b, solve(i,str, mp,j+1));

}
    int maxUniqueSplit(string s) {
        unordered_map<string,int>mp;
        
        return solve( 0, s, mp,0);
    }
};
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