<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 pal(string s, int st, int en){
        for (int i = st; i < (en+1+st)/2; i++) if (s[i] != s[st + en - i]) return false;
        return true;
    }

    vector<vector<string>> comb(vector<string>& k, string s, int st){
        vector<vector<string>> ans;
        if (st == s.size()) ans.push_back(k);
        for(int i = st; i < s.size(); i++){
            if (pal(s, st, i)) {
                k.push_back(s.substr(st, i-st+1));
                vector<vector<string>> b = comb(k, s, i+1);
                k.pop_back();
                ans.insert(ans.end(), b.begin(), b.end());
            }
        }
        return ans;
    }

    vector<vector<string>> partition(string s) {
        vector<string> a;
        return comb(a, s, 0);
    }
};
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