<p>Given a string of digits <code>s</code>, return <em>the number of <strong>palindromic subsequences</strong> of</em> <code>s</code><em> having length </em><code>5</code>. Since the answer may be very large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p><strong>Note:</strong></p>

<ul>
	<li>A string is <strong>palindromic</strong> if it reads the same forward and backward.</li>
	<li>A <strong>subsequence</strong> is a string that can be derived from another string by deleting some or no characters without changing the order of the remaining characters.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;103301&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> 
There are 6 possible subsequences of length 5: &quot;10330&quot;,&quot;10331&quot;,&quot;10301&quot;,&quot;10301&quot;,&quot;13301&quot;,&quot;03301&quot;. 
Two of them (both equal to &quot;10301&quot;) are palindromic.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;0000000&quot;
<strong>Output:</strong> 21
<strong>Explanation:</strong> All 21 subsequences are &quot;00000&quot;, which is palindromic.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;9999900000&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> The only two palindromic subsequences are &quot;99999&quot; and &quot;00000&quot;.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>4</sup></code></li>
	<li><code>s</code> consists of digits.</li>
</ul>

from collections import defaultdict
class Solution:
    def countPalindromes(self, s: str) -> int:
        dp_pre = self.calc_prefixes(s)
        dp_suf = self.calc_suffixes(s)
        mod = 10**9 + 7
        cnts = 0
        for i in range(2, len(s)-2):
            for a in range(10):
                for b in range(10):
                    cnts += dp_pre[(i-1, a, b)] * dp_suf[(i+1, b, a)]
                    cnts %= mod
        return cnts
        
    
    def calc_prefixes(self, s):
        cnts = defaultdict(int)
        # [i][p][c]: the count of pc when the last idx is i and the current number is c.
        dp_pre = defaultdict(int)
        for i, letter in enumerate(s):
            if i > 0:
                for a in range(10):
                    for b in range(10):
                        dp_pre[(i, a, b)] = dp_pre[(i-1, a, b)]
            c = int(letter)
            for p in range(10):
                dp_pre[(i, p, c)] += cnts[p]
            cnts[c] += 1
        return dp_pre


    def calc_suffixes(self, s):
        cnts = defaultdict(int)
        # [i][c][a]: the count of ca when the first idx is i and the current number is c.
        dp_suf = defaultdict(int)
        for i in range(len(s)-1, -1, -1):
            if i < len(s) - 1:
                for a in range(10):
                    for b in range(10):
                        dp_suf[(i, a, b)] = dp_suf[(i+1, a, b)]
            c = int(s[i])
            for a in range(10):
                dp_suf[(i, c, a)] += cnts[a]
            cnts[c] += 1
        return dp_suf
        
        
            
271400