<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>Return the<strong> largest</strong> integer having <code>n</code> digits (as a string) that is <strong>k-palindromic</strong>.</p>

<p><strong>Note</strong> that the integer must <strong>not</strong> have leading zeros.</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">&quot;595&quot;</span></p>

<p><strong>Explanation:</strong></p>

<p>595 is the largest k-palindromic integer with 3 digits.</p>
</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">&quot;8&quot;</span></p>

<p><strong>Explanation:</strong></p>

<p>4 and 8 are the only k-palindromic integers with 1 digit.</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">&quot;89898&quot;</span></p>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= k &lt;= 9</code></li>
</ul>

from functools import cache

class Solution:
    def largestPalindrome(self, n: int, k: int) -> str:
        nn = (n + 1) // 2
        kmod = [[0] * 10 for _ in range(n)]
        num = 1

        for i in range(n):
            for j in range(10):
                kmod[i][j] = num * j % k
            num = num * 10 % k

        val = [''] * nn
        
        @cache
        def solve(i, mod):
            if i >= nn:
                return mod == 0
            
            for dig in range(9, -1, -1):
                f = kmod[i][dig]
                s = kmod[n - i - 1][dig]
                
                if i == n - i - 1:
                    newmod = (mod + f) % k
                else:
                    newmod = (mod + f + s) % k

                if solve(i + 1, newmod):
                    val[i] = chr(dig + ord('0'))
                    return True

            return False

        solve(0, 0)
        t = ''.join(val)
        mx = t + t[::-1] if n % 2 == 0 else t + t[:-1][::-1]

        return mx

157878