<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:
    # https://leetcode.com/problems/palindrome-partitioning-iv/solutions/1042964/python3-dp
    def checkPartitioning(self, s: str) -> bool:
        mp = {}
        for i in range(2*len(s)-1): 
            lo, hi = i//2, (i+1)//2
            while 0 <= lo <= hi < len(s) and s[lo] == s[hi]: 
                mp.setdefault(lo, set()).add(hi)
                lo -= 1
                hi += 1
        
        @lru_cache(None)
        def fn(i, k): 
            """Return True if s[i:] can be split into k palindromic substrings."""
            if k < 0: return False 
            if i == len(s): return k == 0
            return any(fn(ii+1, k-1) for ii in mp[i])
        
        return fn(0, 3)
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