<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:
        
        
        def count_pairs(a,b,inner_range):
            ans = defaultdict(int)
            
            cnt_a = 0
            cnt = 0
            
            for idx in inner_range:
                l = s[idx]
                if l == b:
                    cnt += cnt_a
                if l == a:
                    cnt_a +=1

                ans[idx] = cnt
                
            
            return ans
        
        
        uniq_digit = set(s)
        
        counter_pair = dict()
        counter_pair_reverse = dict()
        
        for a in uniq_digit:
            for b in uniq_digit:
                counter_pair[(a,b)] = count_pairs(a,b,range(len(s)))
                counter_pair_reverse[(b,a)] = count_pairs(a,b,range(len(s)-1,-1,-1))
                
        
        cnt = 0
        mod = 10**9+7
        
        for idx in range(1,len(s)-1):
            for a in uniq_digit:
                for b in uniq_digit:
                    left = counter_pair[(a,b)][idx-1]
                    right = counter_pair_reverse[(b,a)][idx+1]
                    cnt += left*right
                    cnt %= mod
            
        return cnt
    
        
                
                
        
142400