<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:
    bool isPalin(string s){
        int start = 0;
        int end = s.size()-1;

        while(start < end){
            if(s[start] != s[end]){
                return false;
            }
            start++;
            end--;
        }

        return true;
    }

    bool check(vector<string> &temp){
        for(int i=0;i<temp.size();++i){
            if(!isPalin(temp[i])){
                return false;
            }
        }

        return true;
    }

    void f(vector<vector<string>> &ans, vector<string> &temp, string s, int index){
        if(index == s.size()){
            if(check(temp)){
                ans.push_back(temp);
            }
            return;
        }

        for(int i=index;i<s.size();++i){
            string substr = s.substr(index,i-index+1);
            temp.push_back(substr);
            f(ans,temp,s,i+1);
            temp.pop_back();
        }
    }

    vector<vector<string>> partition(string s) {
        vector<vector<string>> ans;
        vector<string> temp;
        f(ans,temp,s,0);
        return ans;
    }
};
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