<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:
        #Ignore string less than 5 characters (cannot have a 5 length substring)
        if len(S) < 5:
            return 0
        
        #Returns the running pair counts for each index
        def get_pairs(s):
            seen_cnt = {str(num):0 for num in range(10)}
            seen_cnt[s[0]] = 1 #We have seen the first character since we start the loop at 1
            pair_cnts = defaultdict(int)
            res = [None] #Filler empty dict (index = 0 / end index)
            
            #Getting running pairs
            for idx in range(1, len(s) - 1):
                res.append(pair_cnts.copy()) #Append running pair counts
                for num in seen_cnt.keys():
                    pair_cnts[(num, s[idx])] += seen_cnt[num]
                seen_cnt[s[idx]] += 1
                
            #Filler empty dict (index = 0 / end index)
            res.append(None)
            
            return res
        
        res = 0
        #Getting post and pre pair counts
        pre, post = get_pairs(S), get_pairs(S[::-1])[::-1]
        
        #Check all possible middle characters -> S[2, len(S) - 2)
        for idx in range(2, len(S) - 2):
            for key, val in pre[idx].items():
                if key in post[idx]:
                    res += post[idx][key] * val #Total pairs per key
                    res %= 1000000007
                    
        return res             
115200