<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>> ans;
    void fun(string s,int i,int prev,vector<string> res){
    if(i==s.size()){
        res.push_back(s.substr(prev,i-prev));
        bool checker=true;
        for(int k=0;k<res.size();k++){
            int l=0;
            int r=res[k].size()-1;
            while(l<r){
                if(res[k][l]!=res[k][r]){
                    checker=false;
                    break;
                }else{
                    l++;r--;
                }
            }
            if(!checker){
                break;
            }
        }
        if(checker){
            ans.push_back(res);
        }
        return;
    }

    res.push_back(s.substr(prev,i-prev));
    fun(s,i+1,i,res);
    res.pop_back();
    fun(s,i+1,prev,res);
}
    vector<vector<string>> partition(string s) {
        vector<string> temp;
        fun(s,1,0,temp);
        return ans;
    }
};
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