<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>

mod  = 10**9 + 7
base = 29
powers = [1]
for i in range(2000): powers.append(powers[-1] * base % mod)

class Solution:
    def checkPartitioning(self, s: str) -> bool:
        n  = len(s)
        h1 = [[0] * n for _ in range(n)]
        h2 = [[0] * n for _ in range(n)]

        for j in range(n):            
            vj = ord(s[j]) - 97
            for i in range(j + 1):
                if i < j:
                    h1[i][j] = (h1[i][j - 1] * base + vj) % mod
                    h2[i][j] = (h2[i][j - 1] + vj * powers[j - i]) % mod
                else:    
                    h1[i][j] = h2[i][j] = vj

        def is_pal(i, j):
            if i  > j: return False
            if i == j: return True       
            m1, m2 = (i + j) // 2, (i + j + 1) // 2
            return h1[i][m1] == h2[m2][j]

        for i in range(n - 2):
            if is_pal(0, i):
                for j in range(i + 1, n - 1):
                    if is_pal(i + 1, j) and is_pal(j + 1, n - 1): return True
        return False
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