<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>

class Solution:
    def countPalindromes(self, s: str) -> int:
        mod = 10 ** 9 + 7

        memo = [[[[-1 for _ in range(6)] for _ in range(11)] for _ in range(11)] for _ in range(len(s))]
        
        def dp(index, a, b, pos):
            if pos > 5:
                return 1

            if index == len(s):
                return 0

            if memo[index][a][b][pos] != -1:
                return memo[index][a][b][pos]

            curr = ord(s[index]) - ord('0')

            result  = dp(index + 1, a, b, pos) % mod

            if pos == 1:
                result = (result + dp(index + 1, curr, b, pos + 1)) % mod
            elif pos == 2:
                result = (result + dp(index + 1, a, curr, pos + 1)) % mod
            elif pos  == 3:
                result = (result + dp(index + 1, a, b, pos + 1)) % mod
            elif pos == 4:
                if curr == b:
                    result = (result + dp(index + 1, a, b, pos + 1)) % mod
            else:
                if curr == a:
                    result = (result + dp(index + 1, a, b, pos + 1)) % mod
            
            memo[index][a][b][pos] = result
            
            return result

        return dp(0, 10, 10, 1)
320800