<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:
        seen = set()
        
        def f(idx, num):
            if idx == (n + 1) // 2:
                mid = (n + 1) // 2
                if n & 1:
                    st = num + num[:-1][::-1]
                else:
                    st = num + num[::-1]                                    
                
                test_num = int(st)
                if test_num % k != 0:
                    return 0
                
                mp_key = "".join(sorted(st))
                if mp_key in seen:                    
                    return 0
                seen.add(mp_key)
                
                # all perms of this palindrome can now be added?
                # remove the dupes as well (n!/a!*b!*c!)
                cnt = Counter(st)
                perm = factorial(n)                
                for c in cnt.values():
                    perm //= factorial(c)
                                        
                # have to account for zeroes - they cannot be leading 0s before arrangement
                # had to look up formulat here: https://math.stackexchange.com/questions/2799149/number-of-permutations-without-leading-zeroes
                # subtract zero counts
                if cnt["0"] > 0:
                    top = factorial(n-1)
                    for key, c in cnt.items():
                        if key == "0":
                            top //= factorial(c - 1)
                        else:
                            top //= factorial(c)                            
                else:
                    top = 0
                    
                # there are still some numbers causing issues, 24042 and 42024
                # just take ascending from prior
                
                return perm - top if test_num % k == 0 else 0                                        
            
            total = 0
            for d in range(10):
                if idx == 0 and d == 0: continue                
                #if len(num) > 0 and idx < (n+1)//2 - 1 and int(num[-1]) > d: continue
                total += f(idx + 1, num + str(d))
                
            return total
        
        return f(0, "")
                
                    
                    
24354