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
function partition(s: string): string[][] {
    const n = s.length;
    const f: boolean[][] = Array.from({ length: n }, () => Array(n).fill(true));
    for (let i = n - 1; i >= 0; --i) {
        for (let j = i + 1; j < n; ++j) {
            f[i][j] = s[i] === s[j] && f[i + 1][j - 1];
        }
    }
    const ans: string[][] = [];
    const t: string[] = [];
    const dfs = (i: number) => {
        if (i === n) {
            ans.push(t.slice());
            return;
        }
        for (let j = i; j < n; ++j) {
            if (f[i][j]) {
                t.push(s.slice(i, j + 1));
                dfs(j + 1);
                t.pop();
            }
        }
    };
    dfs(0);
    return ans;
}
