<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:
    string s;
    int n;
    vector<vector<string>> ans;
    bool isPalindrome(int l,int r){
        while(l<r){
            if(s[l]!=s[r]) return 0;
            l++;
            r--;
        }
        return 1;
    }
    void rec(int x, vector<string> v){
        if(x==n)
        {
            ans.push_back(v);
            return;
        }
        for(int r=x;r<n;r++)
        {
            if(isPalindrome(x,r))
            {
                vector<string> temp = v;
                string res = s.substr(x,r-x+1);
                temp.push_back(res);
                rec(r+1, temp);
            }
        }
    }
    vector<vector<string>> partition(string ss) {
        s=ss;
        n=s.length();
        vector<string> v;
        rec(0,v);
        return ans;
    }
};
241281