<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 {
    private:
bool isPalindrome(string&x){
        for(int i=0;i<x.size()/2;i++){
            if(x[i]!=x[x.size()-1-i]) return false;
        }
        return true;
    }
    void helper( vector<vector<string>>&res,string&s,vector<string>z,string x="",int i=0){
      if(i==s.size()){
        if(isPalindrome(x)){
            z.push_back(x);
            res.push_back(z);    
        }
        return ;
      }
      if(i){
        if(isPalindrome(x))
        z.push_back(x);
        else return;
        }
        x="";
      for(int k=i;k<s.size();k++){
        x+=s[k];
        helper(res,s,z,x,k+1);
      }
    }
    
public:
    vector<vector<string>> partition(string s) {
         vector<vector<string>>res;
    helper(res,s,vector<string>());
    return res;
    }
};
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