<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>

factorials = [1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800]

def get_num_perm(num):
    digit_cnts = {}
    for i in num:
        if i in digit_cnts:
            digit_cnts[i] += 1
        else:
            digit_cnts[i] = 1
    out = 0
    for firstdig in digit_cnts:
        if firstdig != "0":
            nume = factorials[len(num)-1]
            den = 1
            for i in digit_cnts:
                if i==firstdig:
                    den *= factorials[(digit_cnts[i]-1)]
                else:
                    den *= factorials[digit_cnts[i]]
            out += nume // den
    return out

class Solution:
    def countGoodIntegers(self, n: int, k: int) -> int:
        cnt = 0
        d = {}

        if n%2==0:
            for i in range(10**(n//2)):
                curr = int(str(i) + str(i)[::-1])
                if len(str(curr)) < n:
                    continue
                if ''.join(sorted(str(curr))) in d:
                    continue
                if curr % k == 0:
                    cnt += get_num_perm(str(curr))
                    d[''.join(sorted(str(curr)))] = 1
        
        if n%2==1:
            for i in range(10**(n//2)):
                for i2 in range(10):
                    if n==1:
                        curr = i2
                    else:
                        curr = int(str(i) + str(i2) + str(i)[::-1])
                    if len(str(curr)) < n:
                        continue
                    if ''.join(sorted(str(curr))) in d:
                        continue
                    if curr % k == 0:
                        cnt += get_num_perm(str(curr))
                        d[''.join(sorted(str(curr)))] = 1
        return cnt
17431