<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>

def countUniquePerms(num: str, factorialTable, recurse = True):
    freqs = Counter(num)
    totalPerms = factorialTable[len(num)]
    denom = 1
    for freq in freqs.values():
        denom *= factorialTable[freq]
    totalPerms //= denom
    if freqs['0'] > 0 and recurse:
        idx = num.index('0')
        num1 = "".join([num[i] for i in range(len(num)) if i != idx])
        eliminate = countUniquePerms(num1, factorialTable, False)
        totalPerms -= eliminate


    return totalPerms
class Solution:
    def countGoodIntegers(self, n: int, k: int) -> int:
        if n== 1:
            return sum([1 if i %k== 0 else 0 for i in range(1, 10)])
        mid = (n-1) // 2
        count = 0
        factorialTable= [1,1]
        seen = set()
        for i in range(2, n+1):
            factorialTable.append(factorialTable[-1]*i)
        def dfs(number, idx):
            nonlocal count
            if idx > mid:
                prefix = number if n % 2 == 0 else number[:-1]
                newNumber = number + prefix[::-1]
                if int(newNumber) % k == 0:
                    key = "".join(sorted(newNumber))

                    if key in seen:
                        return
                    seen.add(key)
                    count += countUniquePerms(newNumber, factorialTable)
                return
            s = 0 if idx > 0 else 1
            
            for i in range(s, 10):
                dfs(number+str(i), idx+1)
        dfs("", 0)
        return count



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