<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:
        from math import factorial

        total_count = 0

        # Precompute factorials up to n
        factorials = [1] * (n + 1)
        for i in range(1, n + 1):
            factorials[i] = factorials[i - 1] * i

        processed_counts = set()

        # Generate palindromic numbers of length n without leading zeros
        if n == 1:
            start = 1
            end = 9
        else:
            half_length = (n + 1) // 2
            start = 10 ** (half_length - 1)
            end = 10 ** half_length - 1

        for first_half in range(start, end + 1):
            s_str = str(first_half)
            if s_str[0] == '0':
                continue  # Skip palindromes starting with zero
            if n % 2 == 0:
                pal_str = s_str + s_str[::-1]
            else:
                pal_str = s_str + s_str[-2::-1]
            if len(pal_str) != n:
                continue
            pal = int(pal_str)
            if pal % k != 0:
                continue

            # Count the digits
            counts = [0] * 10
            for ch in pal_str:
                counts[int(ch)] += 1

            counts_tuple = tuple(counts)
            if counts_tuple in processed_counts:
                continue
            processed_counts.add(counts_tuple)

            numerator = factorials[n]
            denominator = 1
            for count in counts:
                denominator *= factorials[count]
            total_permutations = numerator // denominator

            # Subtract permutations starting with zero if zero is among the digits
            if counts[0] > 0:
                counts[0] -= 1
                numerator_zero = factorials[n - 1]
                denominator_zero = 1
                for count in counts:
                    denominator_zero *= factorials[count]
                permutations_starting_with_zero = numerator_zero // denominator_zero
                counts[0] += 1
            else:
                permutations_starting_with_zero = 0

            valid_permutations = total_permutations - permutations_starting_with_zero
            total_count += valid_permutations

        return total_count

18230