<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 {
    vector<vector<string>> res;
    string str;
bool isPalindrome(string s) {

    if (s == "")
        return false;

    if(s.size() == 1)
        return true;

    for (int i = 0; i < s.size() / 2; ++i) {
        if (s[i] != s[s.size() - i - 1])
            return false;
    }

    return true;
}
    void dfs(vector<vector<string>>& res , string s,int st, vector<string> subset){

        if(isPalindrome(s)){
            subset.push_back(s);
        }else if(s != ""){
            return;
        }

        if(str.size() == st)
            res.push_back(subset);

        for(int i = 1; st + i <= str.size()  ;++i){
            string newStr = str.substr(st, i);
            dfs(res,newStr,st + i,subset);
        }
    }
public:
    vector<vector<string>> partition(string s) {
        this->str = s;
        dfs(res,"",0,{});
        return res;
    }
};
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