<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) {
        unordered_set<string> usedSubstrings;
        return backtrack(s, 0, usedSubstrings);
    }

private:
    // Backtracking function to split the string and find maximum number of unique substrings
    int backtrack(const string& s, int start, unordered_set<string>& usedSubstrings) {
        if (start == s.length()) {
            return 0;  // No more characters to split, return 0
        }

        int maxCount = 0;

        // Try all possible substrings starting from index `start`
        for (int end = start + 1; end <= s.length(); end++) {
            string currentSubstring = s.substr(start, end - start);  // Get the current substring

            // If the substring has not been used before, use it and continue
            if (usedSubstrings.find(currentSubstring) == usedSubstrings.end()) {
                usedSubstrings.insert(currentSubstring);  // Add substring to the set
                // Recursively split the remaining string and count the unique substrings
                maxCount = max(maxCount, 1 + backtrack(s, end, usedSubstrings));
                usedSubstrings.erase(currentSubstring);  // Backtrack by removing the substring
            }
        }

        return maxCount;  // Return the maximum count of unique substrings
    }
};

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