Palindrome Partitioning
<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>
String; Dynamic Programming; Backtracking
class Solution {
public:
    vector<vector<string>> partition(string s) {
        int n = s.size();
        bool f[n][n];
        memset(f, true, sizeof(f));
        for (int i = n - 1; i >= 0; --i) {
            for (int j = i + 1; j < n; ++j) {
                f[i][j] = s[i] == s[j] && f[i + 1][j - 1];
            }
        }
        vector<vector<string>> ans;
        vector<string> t;
        auto dfs = [&](this auto&& dfs, int i) -> void {
            if (i == n) {
                ans.emplace_back(t);
                return;
            }
            for (int j = i; j < n; ++j) {
                if (f[i][j]) {
                    t.emplace_back(s.substr(i, j - i + 1));
                    dfs(j + 1);
                    t.pop_back();
                }
            }
        };
        dfs(0);
        return ans;
    }
};
