<p>Given a string <code>s</code>, partition <code>s</code> such that every <span data-keyword="substring-nonempty">substring</span> of the partition is a <span data-keyword="palindrome-string"><strong>palindrome</strong></span>. Return <em>all possible palindrome partitioning of </em><code>s</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> s = "aab"
<strong>Output:</strong> [["a","a","b"],["aa","b"]]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> s = "a"
<strong>Output:</strong> [["a"]]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 16</code></li>
	<li><code>s</code> contains only lowercase English letters.</li>
</ul>

class Solution {
    void rec(
        int i, 
        string s, 
        vector<string>& current,
        string undividedStr,
        vector<vector<string>>& res
    ) {
        if (i == s.size()) {
            res.push_back(current);
            return;
        }

        undividedStr += s[i];
        cout << "i: "<< i << " " << "undivided: "<< undividedStr << endl;
        if (isPalindrome(undividedStr)) {
            current.push_back(undividedStr);
            rec(i+1, s, current, "", res);
            current.pop_back();
        }
        
        if (i != s.size()-1) {
            rec(i+1, s, current, undividedStr, res);
        }
    }

    // [a, a, b]
    

    bool isPalindrome(string s) {
        // [a, b, c, d,]
        for (int i = 0; i < s.size()/2; i++) {
            if (s[i] != s[s.size() - i - 1]) {
                return false;
            }
        }

        return true;
    }

public:
    vector<vector<string>> partition(string s) {
        vector<vector<string>> res;
        vector<string> current;
        rec(0, s, current, "", res);

        return res;
    }
};
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