<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) 
    {
        int n = s.size();

        auto is_palindrome =[&]( string str )
        {
            string rev_str = str;
            reverse( rev_str.begin() , rev_str.end() );
            return str == rev_str;
        };
        
        vector<vector<string>> res;
        vector<string> temp;

        function<void(int,string)> f =[&](int i , string str)
        {
            if( i == n ) 
            {
                if( str.empty() || is_palindrome(str) )
                {
                    if (!str.empty()) temp.push_back( str );
                    res.push_back( temp );
                    if (!str.empty()) temp.pop_back();
                }
                return;
            }

            f(i + 1 , str + s[i]);

            if( !str.empty() && is_palindrome(str) )
            {
                temp.push_back( str );
                f(i + 1 , string(1, s[i]));
                temp.pop_back();
            }
            
        };

        f(0 , "");
        return res;
    }
};

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