<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:
        BASE = 10

        def to_counts_hash(counter: collections.Counter[str]) -> int:
            h = 0
            for d in range(BASE):
                h = h * BASE + counter[str(d)]
            return h

        appeared_representations = {}

        if n == 1:
            return len({num for num in range(1, BASE) if num % k == 0})

        half_len = n // 2
        if n % 2 == 0:
            for t in range(1, BASE ** half_len):
                tail = str(t).zfill(half_len)
                combined = tail[::-1] + tail
                if combined[0] == '0':
                    continue
                result_num = int(combined)
                if result_num % k != 0:
                    continue

                counter = collections.Counter(combined)
                h = to_counts_hash(counter)
                appeared_representations[h] = counter
        else:
            for i in range(BASE):
                for t in range(1, BASE ** half_len):
                    tail = str(t).zfill(half_len)
                    combined = tail[::-1] + str(i) + tail
                    if combined[0] == '0':
                        continue
                    result_num = int(combined)
                    if result_num % k != 0:
                        continue

                    counter = collections.Counter(combined)
                    h = to_counts_hash(counter)
                    appeared_representations[h] = counter

        answer = 0
        for counter in appeared_representations.values():
            comb = math.factorial(n)
            for count in counter.values():
                comb //= math.factorial(count)
            answer += comb

            if counter['0'] > 0:
                reduction = math.factorial(n - 1)
                for d, count in counter.items():
                    if d == '0':
                        reduction //= math.factorial(count - 1)
                    else:
                        reduction //= math.factorial(count)
                answer -= reduction

        return answer

19295