<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 {
public:
    vector<vector<string>> partition(string s) {
        vector<vector<string>> combs;
        vector<string> comb;
        backtrack(s, combs, comb, "", 0);
        return combs;
    }

    void backtrack(string& s, vector<vector<string>>& combs, vector<string>& comb, string comb2, int index) {
        if (index == s.length()) {
            combs.push_back(comb);
            return;
        }

        for(int i = index; i < s.length(); i++) {
            string currentSub = s.substr(index, i - index + 1);
            string temp = currentSub;
            reverse(temp.begin(), temp.end());
            if(temp == currentSub) {
                comb.push_back(currentSub);
                backtrack(s, combs, comb, comb2 + currentSub, i + 1);
                comb.pop_back();
            }
        }
    }
};
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