<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:
        # before[i] = dictionary containing the number of pairs of each possible combo. Keys are "00", "01" etc. Pairs that end before or at ith index
        # after[i] = dictionary containing the number of pairs of each possible combo. Keys are "00", "01" etc. Pairs that start after or on the ith index

        L = len(s)

        from collections import defaultdict

        before = [defaultdict(int) for _ in range(L)]
        after = [defaultdict(int) for _ in range(L)]

        freq = defaultdict(int) # frequency of current characters encountered

        freq[s[0]] += 1

        for i in range(1, L):
            cur_char = s[i]
            for k,v in freq.items():
                cur_pair = f"{k}{cur_char}"
                before[i][cur_pair] = v

            for k,v in before[i-1].items():
                before[i][k] += v

            freq[cur_char] += 1

         
        freq = defaultdict(int) # frequency of current characters encountered

        freq[s[-1]] += 1

        for i in range(L - 2, -1, -1):
            cur_char = s[i]
            for k,v in freq.items():
                cur_pair = f"{cur_char}{k}"
                after[i][cur_pair] = v

            for k,v in after[i+1].items():
                after[i][k] += v

            freq[cur_char] += 1

        total = 0
        for middle in range(2, L-2):
            for start in range(0, 10):
                for end in range(0, 10):
                    cur_pair = f"{start}{end}"
                    rev_pair = f"{end}{start}"
                    total = (total + before[middle - 1][cur_pair] * after[middle + 1][rev_pair]) % (10**9 + 7)

        return total




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