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

    bool allPalindrome(vector<string> &v) {
        for(string &x : v) {
            if(!palindrome(x)) {
                return false;
            }
        }
        return true;
    }

    void dfs(vector<string> v, vector<vector<string>>& ans) {
        if(allPalindrome(v)) {
            ans.push_back(v);
        }
        string partition = v[v.size() - 1];
        v.pop_back();
        for(int i = 1; i < partition.size(); i++) {
            v.push_back(partition.substr(0, i));
            v.push_back(partition.substr(i, partition.size() - i));
            dfs(v, ans);
            v.pop_back();
            v.pop_back();
        }
    }
    
    vector<vector<string>> partition(string s) {
        vector<vector<string>> ans;
        vector<string> cur;
        cur.push_back(s);
        dfs(cur, ans);
        return ans;
    }
};
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