<p>You are given two <strong>positive</strong> integers <code>n</code> and <code>k</code>.</p>

<p>An integer <code>x</code> is called <strong>k-palindromic</strong> if:</p>

<ul>
	<li><code>x</code> is a <span data-keyword="palindrome-integer">palindrome</span>.</li>
	<li><code>x</code> is divisible by <code>k</code>.</li>
</ul>

<p>An integer is called <strong>good</strong> if its digits can be <em>rearranged</em> to form a <strong>k-palindromic</strong> integer. For example, for <code>k = 2</code>, 2020 can be rearranged to form the <em>k-palindromic</em> integer 2002, whereas 1010 cannot be rearranged to form a <em>k-palindromic</em> integer.</p>

<p>Return the count of <strong>good</strong> integers containing <code>n</code> digits.</p>

<p><strong>Note</strong> that <em>any</em> integer must <strong>not</strong> have leading zeros, <strong>neither</strong> before <strong>nor</strong> after rearrangement. For example, 1010 <em>cannot</em> be rearranged to form 101.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">n = 3, k = 5</span></p>

<p><strong>Output:</strong> <span class="example-io">27</span></p>

<p><strong>Explanation:</strong></p>

<p><em>Some</em> of the good integers are:</p>

<ul>
	<li>551 because it can be rearranged to form 515.</li>
	<li>525 because it is already k-palindromic.</li>
</ul>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">n = 1, k = 4</span></p>

<p><strong>Output:</strong> <span class="example-io">2</span></p>

<p><strong>Explanation:</strong></p>

<p>The two good integers are 4 and 8.</p>
</div>

<p><strong class="example">Example 3:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">n = 5, k = 6</span></p>

<p><strong>Output:</strong> <span class="example-io">2468</span></p>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 10</code></li>
	<li><code>1 &lt;= k &lt;= 9</code></li>
</ul>

from itertools import permutations
from math import factorial
class Solution:
    def check(self, s):
        return s == s[::-1]
    def countGoodIntegers(self, n: int, k: int) -> int:
        
        hm = {'1','2', '3','4','5','6','7','8','9'}
        i = 0
        for i in range((n-2)//2):
            l = list(hm)
            hm = set()
            for j in range(10):
                for m in l:
                    m += str(j)
                    hm.add(m)
        
        if (n-2)% 2 and n > 1:
            l = list(hm)
            hm = set()
            for i in l:
                for j in range(10):
                    a = i + str(j) + i[::-1]
                    if int(a) % k ==0:
                        p = list(a)
                        p.sort()
                        hm.add(''.join(p))
        elif (n-2)%2==0 and n > 0:
            l = list(hm)
            hm = set()
            for i in l:
                a = i + i[::-1]
                if int(a) % k == 0:
                    p = list(a)
                    p.sort()
                    hm.add(''.join(p))
        # print(hm)
        if n == 1:
            hm = set()
            for p in range(1,10):
                if p%k==0:
                    hm.add(str(p))
        print(hm)
        ans = 0
        for i in hm:
            a = 1
            cnt = [0]*10
            for j in i:
                cnt[int(j)] += 1
                
            a = factorial(n)
            for p in range(0,10):
                a //= factorial(cnt[p])
                
            '''for p in range(1, 10):
                a //= factorial(cnt[p])'''
            if '0' in i:
                cnt[0] -= 1
                b = factorial(n-1)
                for p in range(0,10):
                    b //= factorial(cnt[p])
                a -= b
            ans += a
            print(a, i)
            
        return ans
28880