<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:
        '''
        3 <= s.length <= 2000
        
        we have all possible palindromes
        run dfs from 0, and count = 3
        '''
        N = len(s)
        arr = set()
        
        for i in range(N):
            l, r = i, i
            
            while l >= 0 and r < N and s[l] == s[r]:
                arr.add((l, r))
                l -= 1
                r += 1
                
        for i in range(N):
            l, r = i, i + 1
            
            while l >= 0 and r < N and s[l] == s[r]:
                arr.add((l, r))
                l -= 1
                r += 1
                
                
                
       
        lookup = defaultdict(list)
        
        for start, end in arr:
            lookup[start].append(end)
        
        
        
        @cache
        def calc(index, amount):
            if index == N:
                return bool(amount == 0)
            
            if amount == 0:
                return False
            
            
            best = False
            
            for v in lookup[index]:
                best |= calc(v + 1, amount - 1)
                
            return best
        
        
        
        return calc(0, 3)
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