<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 math import factorial
from collections import Counter

class Solution:
    def countGoodIntegers(self, n: int, k: int) -> int:

        def generateNums(m):  # generates all nums of length m
            if m <= 0:
                return []

            start = 10**(m - 1)
            end = 10**m

            return list(range(start, end))

        count = 0
        if n == 1:
            for i in range(1, 10):
                if i % k == 0:
                    count += 1
            return count

        palindromes = set()

        if n % 2 == 0: #generate all even palindromes
            m = n // 2  
            nums = generateNums(m)
            for num in nums:
                palindrome = str(num) + str(num)[::-1]
                if int(palindrome) % k == 0:
                    cleaned_palindrome = ''.join(sorted(palindrome))
                    if cleaned_palindrome:  # Only add non-empty strings
                        palindromes.add(cleaned_palindrome)

        if n % 2 != 0: #generate all odd palindromes
            m = n // 2
            nums = generateNums(m)
            for num in nums:
                for i in range(0, 10):
                    palindrome = str(num) + str(i) + str(num)[::-1]
                    if int(palindrome) % k == 0:
                        cleaned_palindrome = ''.join(sorted(palindrome))
                        if cleaned_palindrome:  # Only add non-empty strings
                            palindromes.add(cleaned_palindrome)
        #print(palindromes)
        def countPermutations(s): #function to count all permutations of a string s that dont have leading 0s
            char_count = Counter(s)
            n = len(s)
            
            # Total permutations
            total_permutations = factorial(n)
            for count in char_count.values():
                total_permutations //= factorial(count)
            
            # If there's no '0', all permutations are valid
            if '0' not in char_count:
                return total_permutations
            
            # Calculate permutations that start with '0'
            char_count['0'] -= 1
            leading_zero_permutations = factorial(n - 1)
            for count in char_count.values():
                leading_zero_permutations //= factorial(count)
            
            # Subtract permutations with leading '0' from total permutations
            return total_permutations - leading_zero_permutations

        for pal in palindromes:
            count += countPermutations(pal)

        return count
19828