<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:
        n=len(s)
        palin=[[False]*n for i in range(n)]
        for i in range(n):
            palin[i][i]=True
        for l in range(2,n):
            for i in range(n-l+1):
                j=i+l-1
                if s[i]==s[j]:
                    if l==2:
                        palin[i][j]=True
                    elif palin[i+1][j-1]:
                        palin[i][j]=True
                
        def backtrack(i,k):
            if i>=n:
                return False
            if k==1:
                return palin[i][n-1]
            res=False
            for j in range(i,n):
                res|=(backtrack(j+1,k-1) and palin[i][j])
            return res

            
        return backtrack(0,3)
            
        
#         def palin(k):  TIME LIMIT EXCEEDED
#             n=len(k)
#             l=0
#             r=n-1
#             while l<=r:
#                 if k[l]!=k[r]:
#                     return False
#                 l+=1
#                 r-=1
#             return True
            
#         def backtrack(i,k):
#             if i>=n:
#                 return False
#             if k==1:
#                 return palin(s[i:])
#             res=False
#             for j in range(i,n):
#                 res|=backtrack(j+1,k-1) and palin(s[i:j+1])
#             return res

            
#         return backtrack(0,3)
3186