<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 check(vector<string>& temp) {
        for (auto str : temp) {
            int sz = str.size();
            for (int i = 0; i < sz / 2; i++) {
                if (str[i] != str[sz - 1 - i])
                    return false;
            }
        }
        return true;
    }

    void solve(int i, string curr, vector<string> temp, int n, string& s,
               vector<vector<string>>& ans) {
        if (i == n) {
            // check if all strings in vector is palindrome
            if (curr.size() > 0)
                temp.push_back(curr);
            if (temp.size() > 0 && check(temp)) {
                ans.push_back(temp);
            }
            return;
        }

        // no partition
        solve(i + 1, curr + s[i], temp, n, s, ans);

        // partition
        // if(curr.size()>0) {
        temp.push_back(curr);
        string str = "";
        str += s[i];
        solve(i + 1, str, temp, n, s, ans);
        temp.pop_back();
        // }
    }

    vector<vector<string>> partition(string s) {
        vector<vector<string>> ans;
        vector<string> temp;
        string str = "";
        str += s[0];
        solve(1, str, temp, s.size(), s, ans);
        return ans;
    }
};
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