<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) {
       vector<string>ds;
       vector<vector<string>>ans;
       f(s, 0 ,ds, ans);
       return ans; 
    }
void f( string s, int ind , vector<string> &ds, vector<vector<string>> &ans){
    if(ind == s.size()){
        ans.push_back(ds);
        return;
    }
    for(int i = ind; i<s.size();i++){
    if(isPalindrome(s, ind, i)){
        ds.push_back(s.substr(ind, i-ind+1));
        f(s, i+1,ds, ans);
        ds.pop_back();
    }
    
    }
}


    bool isPalindrome( string s , int start , int end){
        while( start<= end){
            if(s[start++] != s[end--]){
                return false;
            }
        }
        return true;
    }
};
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