<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>

class Solution:
    def checkPartitioning(self, s: str) -> bool:
        parlin_set = set()
        for i in range(len(s)):
            for j in range(i, (len(s))):
                temp = s[i:j+1]
                if temp == temp[::-1]:
                    parlin_set.add((i,j))
        cache = {}
        def dfs(i, partitions) -> bool:
            if i == len(s):
                return partitions == 0
            if partitions < 0: return False
            if (i, partitions) in cache:
                return cache[(i, partitions)]
            temp = []
            for j in range(i, len(s)):
                temp.append(s[j])
                if (i, j) in parlin_set:
                    if dfs(j+1, partitions - 1):
                        cache[(i, partitions)] = True
                        return cache[(i, partitions)]
            cache[(i, partitions)] = False
            return cache[(i, partitions)]
        return dfs(0, 3)
389400