<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 {
private:
    int solve(int ind, string& s, unordered_set<string>& st, string temp){
        if(ind == s.length())
            return 0;
        else{
            temp.push_back(s[ind]);
            int op1 = -1;
            int op2 = -1;
			// if new substring is not present in the set
            if(st.find(temp) == st.end()){
                st.insert(temp);
                op1 = 1 + solve(ind+1, s, st, "");  //since new substring is added we increase the cnt by 1 and create a new substring to check from very next index -----> in other words we pick that string
                st.erase(temp);
            }
			// picked part was done,  its time for not picking the char at that index and moving to the next one
            op2 = solve(ind+1, s, st, temp);
            return max(op1, op2);
        }
    }
public:
    int maxUniqueSplit(string s) {
        // creating unordered_set to keep track which substring is visited
        unordered_set<string> st;
		// recursion function
        return solve(0, s, st, ""); 
    }
};
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