<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 solve(self, idx, s, dp, three_cuts):
        if idx == len(s):
            return three_cuts == 3
        if three_cuts >= 3:
            return False
        if (idx, three_cuts) in dp:
            return dp[(idx, three_cuts)]
        possible = False
        for i in range(idx, len(s)):
            if s[idx: i + 1] == s[idx: i + 1][::-1]:
                possible |= self.solve(i + 1, s, dp, three_cuts + 1)
        dp[(idx, three_cuts)] = possible
        return dp[(idx, three_cuts)]
    
    def checkPartitioning(self, s: str) -> bool:
        return self.solve(0, s, dict(), 0)
        
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