<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:
        pm = [0] * (n + 10)
        p = 1
        kk = k
        for i in range(n + 5):
            pm[i] = p
            p = (p * 10) % kk
        
        res=[-1] * n
        @cache
        def dfs(i,k):
            if n%2==0 and i>=n//2:
                return k==0
            if n%2==1 and i>n//2:
                return k==0
            w=pm[n-1-i] + pm[i]
            if n%2==1 and i==n//2:
                w=pm[i]
            for j in range(9,-1,-1):
                if j==0 and i==0:
                    continue
                r = dfs(i+1, (kk + k - (w*j)%kk) % kk)
                if r:
                    if res[i] != -1:
                        raise Exception()
                    res[i] = j
                    res[n-1-i] = j
                    return True
            return False
        
        dfs(0,kk)
        return ''.join(str(c) for c in res)

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