<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:

    bool isPalindrome(string temp) {
        string str = temp;
        reverse(temp.begin(), temp.end());
        return str == temp;
    }
    void bt(vector<vector<string>>& res, string s, vector<string>& str, int start) {
        if(start == s.size()) {
            res.push_back(str);
            return;
        }
        for(int i = start; i < s.size(); i++) {
            if(isPalindrome(s.substr(start, i-start+1))) {
                str.push_back(s.substr(start, i-start+1));
                bt(res, s, str, i+1);
                str.pop_back();
            }
        }
        
    }
    vector<vector<string>> partition(string s) {
        vector<vector<string>> res;
        vector<string> str;
        bt(res, s, str, 0);
        return res;
    }
};
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