<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 checkpalind(self,s):
        return s==s[::-1]
    def canpalind(self,ind,s,cuts,dp):
        if ind==len(s):
            if cuts==0:
                return True
            return False
        if dp[ind][cuts]!=-1:
            return dp[ind][cuts]
        for i in range(ind,len(s)):
            if self.checkpalind(s[ind:i+1]) and cuts>0:
                if self.canpalind(i+1,s,cuts-1,dp):
                    dp[ind][cuts]=True
                    return dp[ind][cuts]
        dp[ind][cuts]=False
        return dp[ind][cuts]
    def checkPartitioning(self, s: str) -> bool:
        dp=[[-1 for _ in range(4)]for i in range(len(s))]
        return self.canpalind(0,s,3,dp)
        
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