<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 {
private:
    bool isPalindrome(const string& s, int start, int end) {
        while (start < end) {
            if (s[start] != s[end]) return false;
            start++;
            end--;
        }
        return true;
    }

    vector<vector<string>> res;

public:

    void dfs(string s, int start, vector<string>& path) {
        if (start == s.size()) {
            vector<string> toAdd(path);
            res.push_back(toAdd);
            return;
        }
        for (int i = 0; i < s.size() - start; i++) {
            if (!isPalindrome(s, start, start + i)) {
                continue;
            }
            path.push_back(s.substr(start, i + 1));
            dfs(s, start + i + 1, path);
            path.pop_back();
        }
    }

    vector<vector<string>> partition(string s) {
        res.clear();
        vector<string> path {};
        dfs(s, 0, path);
        return res;
    }
};
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