Find the Count of Good Integers
<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>
Hash Table; Math; Combinatorics; Enumeration
class Solution {
    public long countGoodIntegers(int n, int k) {
        long[] fac = new long[n + 1];
        fac[0] = 1;
        for (int i = 1; i <= n; i++) {
            fac[i] = fac[i - 1] * i;
        }

        long ans = 0;
        Set<String> vis = new HashSet<>();
        int base = (int) Math.pow(10, (n - 1) / 2);

        for (int i = base; i < base * 10; i++) {
            String s = String.valueOf(i);
            StringBuilder sb = new StringBuilder(s).reverse();
            s += sb.substring(n % 2);
            if (Long.parseLong(s) % k != 0) {
                continue;
            }

            char[] arr = s.toCharArray();
            Arrays.sort(arr);
            String t = new String(arr);
            if (vis.contains(t)) {
                continue;
            }
            vis.add(t);
            int[] cnt = new int[10];
            for (char c : arr) {
                cnt[c - '0']++;
            }

            long res = (n - cnt[0]) * fac[n - 1];
            for (int x : cnt) {
                res /= fac[x];
            }
            ans += res;
        }

        return ans;
    }
}
