Palindrome Partitioning IV
<p>Given a string <code>s</code>, return <code>true</code> <em>if it is possible to split the string</em> <code>s</code> <em>into three <strong>non-empty</strong> palindromic substrings. Otherwise, return </em><code>false</code>.​​​​​</p>

<p>A string is said to be palindrome if it the same string when reversed.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;abcbdd&quot;
<strong>Output:</strong> true
<strong>Explanation: </strong>&quot;abcbdd&quot; = &quot;a&quot; + &quot;bcb&quot; + &quot;dd&quot;, and all three substrings are palindromes.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;bcbddxy&quot;
<strong>Output:</strong> false
<strong>Explanation: </strong>s cannot be split into 3 palindromes.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>3 &lt;= s.length &lt;= 2000</code></li>
	<li><code>s</code>​​​​​​ consists only of lowercase English letters.</li>
</ul>
String; Dynamic Programming
class Solution {
    public boolean checkPartitioning(String s) {
        int n = s.length();
        boolean[][] f = new boolean[n][n];
        for (var g : f) {
            Arrays.fill(g, true);
        }
        for (int i = n - 1; i >= 0; --i) {
            for (int j = i + 1; j < n; ++j) {
                f[i][j] = s.charAt(i) == s.charAt(j) && (i + 1 == j || f[i + 1][j - 1]);
            }
        }
        for (int i = 0; i < n - 2; ++i) {
            for (int j = i + 1; j < n - 1; ++j) {
                if (f[0][i] && f[i + 1][j] && f[j + 1][n - 1]) {
                    return true;
                }
            }
        }
        return false;
    }
}
