<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>

class Solution:
    def countGoodIntegers(self, n: int, k: int) -> int:
        # Generate all palindromes of length n
        def gen_all_palindromes(n):
            if n == 1:
                # Exclude 0 since it technically has a leading zero
                return [i for i in range(1, 10)]
            
            if n % 2 == 0:
                start = 10**(n//2 - 1)
                end = 10**(n//2)
            else:
                start = 10**(n//2)
                end = 10**(n//2 + 1)

            palindromes = []
            for i in range(start, end):
                half = str(i)

                # All palindromes of even length
                if n % 2 == 0:
                    palindromes.append(int(half + half[::-1]))
                # All palindromes of odd length
                else:
                    palindromes.append(int(half + half[-2::-1]))
            
            return palindromes
        
        # Get all palindromes that are divisible by k
        k_palindromes = [p for p in gen_all_palindromes(n) if p % k == 0]
        print(k_palindromes)

        # A set of tuples, each of length 10, with the count of each digit
        # We know we can rearrange a number into another number if it has the same
        # amount of each digit
        digit_counts = set()

        for p in k_palindromes:
            # Get all counts of each digit
            # Then, we go through each digit and count how many times it appears
            digit_counts.add(tuple(sorted(Counter(str(p)).items())))

        num_good = 0
        fact = math.factorial(n)

        for counts in digit_counts:
            denom = 1

            # We can calculate the number of permuations without actually having to generate them
            # We just need to divide the total number of permutations by the number of times each digit appears
            for _, num in counts:
                denom *= math.factorial(num)

            # Boom! fast and easy
            total_perms = fact // denom
            # Exclude all permutations where they lead with zeroee
            invalid_perms = 0
            # All other permutations are valid, so we just need to cacluate these invalid ones
            if '0' in dict(counts):
                num_zero = dict(counts)['0']
                invalid_perms = (num_zero * math.factorial(n-1)) // denom

            # Add the total number we found plus the number we found with leading zeroes
            num_good += total_perms - invalid_perms
        
        return num_good

26484