<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>

from math import factorial
from collections import Counter

class Solution:
    def countGoodIntegers(self, n: int, k: int) -> int:
    
        fact = [1] * (n + 1)
        for i in range(1, n + 1):
            fact[i] = fact[i - 1] * i

        def generate_palindromes():
            half_len = n // 2
            palindromes = []
         
            def backtrack(pos, half_digits):
                if pos == half_len:
                    if n % 2 == 0:
                        full_num = half_digits + half_digits[::-1]
                    else:
                        for mid_digit in range(10):
                            full_num = half_digits + [mid_digit] + half_digits[::-1]
                            if full_num[0] == 0:
                                continue
                            num = int(''.join(map(str, full_num)))
                            if num % k == 0:
                                palindromes.append(full_num)
                    if n % 2 == 0:
                        if half_digits[0] == 0:
                            return
                        full_num = half_digits + half_digits[::-1]
                        num = int(''.join(map(str, full_num)))
                        if num % k == 0:
                            palindromes.append(full_num)
                    return
                
                start_digit = 1 if pos == 0 else 0
                for digit in range(start_digit, 10):
                    half_digits.append(digit)
                    backtrack(pos + 1, half_digits)
                    half_digits.pop()
            
            backtrack(0, [])
            return palindromes
        
        def count_permutations(digits):
            freq = Counter(digits)
            total = fact[n]
            for count in freq.values():
                total //= fact[count]
           
            if 0 not in freq:
                return total
            else:
                total_perms = 0
                for leading_digit in range(1, 10):
                    if freq[leading_digit] == 0:
                        continue
                    freq[leading_digit] -= 1
                    perms = fact[n - 1]
                    for count in freq.values():
                        perms //= fact[count]
                    total_perms += perms
                    freq[leading_digit] += 1
                return total_perms
        
        palindromes = generate_palindromes()
        result = 0
        unique_digit_configs = set()
        
        for pal in palindromes:
            digits = pal
            digit_tuple = tuple(sorted(digits))
            if digit_tuple in unique_digit_configs:
                continue
            unique_digit_configs.add(digit_tuple)
            permutations_count = count_permutations(digits)
            result += permutations_count
        
        return result

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