<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<vector<string>> out;
        vector<pair<size_t, size_t>> temp;
        backtrack(out, s, 0, temp);
        return out;
    }

    void backtrack(vector<vector<string>>& out, const string& s, size_t start, vector<pair<size_t, size_t>>& temp) {
        if (start > s.size())
            return;
        if (start == s.size()) {
            auto& v = out.emplace_back();
            for (const auto& p: temp) {
                v.push_back(s.substr(p.first, p.second-p.first+1));
            }
            return;
        }
        for (size_t i = start; i < s.size(); i++) {
            if (!palindrome(s, start, i)) continue;

            temp.push_back({start, i});
            backtrack(out, s, i+1, temp);
            temp.pop_back();
        }
    }
    // void backtrack(vector<vector<string>>& out, const string& s, size_t start, vector<string>& temp) {
    //     if (start > s.size())
    //         return;
    //     if (start == s.size()) {
    //         out.push_back(temp);
    //         return;
    //     }
    //     for (size_t i = start; i < s.size(); i++) {
    //         if (!palindrome(s, start, i)) continue;

    //         temp.push_back(s.substr(start, i-start+1));
    //         backtrack(out, s, i+1, temp);
    //         temp.pop_back();
    //     }
    // }

    bool palindrome(const string& s, size_t b, size_t e) {
        while (b < e) {
            if (s[b++] != s[e--])
                return false;
        }
        return true;
    }
};
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