<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>

MOD = 10**9+7
class Solution:
    def countPalindromes(self, s: str) -> int:
        n = len(s)
        right = [  [0]*100 for i in range(n)  ]
        count = [0] * 10
        count[int(s[n-1])] = 1
        for r in reversed(range(n-1)):
            x = int(s[r])
            for i in range(100):
                right[r][i] = right[r+1][i]
                if i//10 == x:
                    right[r][i] = (right[r][i] + count[i%10]) % MOD
                # if right[r][i] > 0:
                #     print(i,right[r][i],end=' , ')
            count[x] += 1
            # print()
                

    
        
        left = [0] * 100
        count = [0] * 100
        res = 0
        count[int(s[0])] = 1
        for l in range(1,n-2):
            x = int(s[l])
            for i in range(100):
                res = (res + left[i] * right[l+1][ 10*(i%10) + i//10  ]) % MOD

            for i in range(100):
                if i%10 == x:
                    left[i] = (left[i] + count[i//10]) % MOD
            count[x] += 1
            
        
        return res 









        # right = [None] * (n-1) + [[0]*10 for i in range(10)]
        # n = len(s)
        # count = [0] * 10
        # count[int(s[n-1])] =  1
        
        # for k in range(n-1)
        #     x = int(nums[k])
        #     curr = [ [ right[k+1][i][j] for j in range(10) ]  for i in range(10) ]

        #     for j in range(10):
        #         curr[x][j] = count[j]
            
        #     right[k] = curr
        #     count[x] += 1
        
        # left = [[0]*10 for i in range(10)]
        # count = [0]*10
        # count[int(s[0])] = 1
        # res = 0
        # for k in range(n-2):
        #     x = int(nums[k])
        #     for i in range(10):
        #         for j in range(10):
        #             res += min(left[i][j], right[k+1][j][i])
        #             res%= MOD

        #     for i in range(10):
        #         left[i][x] = count[i]
        #     count[x] += 1
        
            











        
28200