<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:
        
        if len(s) < 5:
            return 0

        ans = 0

        N = len(s)
        dig_occur_l = [[0 for _ in range(10)] for _ in range(N)]
        dig_occur_r = [[0 for _ in range(10)] for _ in range(N)]

        pair_occur_l = [[0 for _ in range(100)] for _ in range(N)]
        pair_occur_r = [[0 for _ in range(100)] for _ in range(N)]

        # lr pass
        for i in range(N):
            if i > 0:
                
                for j in range(10):
                    dig_occur_l[i][j] = dig_occur_l[i-1][j]
            
                for p in range(100):
                    pair_occur_l[i][p] = pair_occur_l[i-1][p] 
                    if p % 10 == int(s[i]):
                        pair_occur_l[i][p] += dig_occur_l[i-1][p//10]

            dig_occur_l[i][int(s[i])] += 1

        # rl pass
        for i in range(N-1, -1, -1):
            if i < N-1:
                for j in range(10):
                    dig_occur_r[i][j] = dig_occur_r[i+1][j]
            
                for p in range(100):
                    pair_occur_r[i][p] = pair_occur_r[i+1][p] 
                    if p // 10 == int(s[i]):
                        pair_occur_r[i][p] += dig_occur_r[i+1][p%10]
                    
                    if i >= 2 and i < N-2:
                        l = pair_occur_l[i-1][p]
                        r = pair_occur_r[i+1][(p%10)*10 + (p//10)]
                        ans = (ans + (l * r)) % 1000000007

            dig_occur_r[i][int(s[i])] += 1
                

        # for i in range(N):
        #     print(f"Left of {i}: {[(p, pair_occur_l[i][p]) for p in range(100) if pair_occur_l[i][p] != 0]}")
        #     print(f"Right of {i}: {[(p, pair_occur_r[i][p]) for p in range(100) if pair_occur_l[i][p] != 0]}")
            

        # for i in range(2, N-2):
        #     for p in range(100):
        #         l = pair_occur_l[i-1][p]
        #         r = pair_occur_r[i+1][(p%10)*10 + (p//10)]
        #         ans = (ans + (l * r)) % 1000000007
        
        return ans 
            



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