<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>

class Solution:
    def largestPalindrome(self, n: int, k: int) -> str:
        if k == 1:
            return '9' * n

        L = (n - 1) // 2
        m = [1] * (n)
        d = 1
        for i in range(n):
            m[i] = d
            d = (d * 10) % k
        m = m[::-1]
        # print(L, m)

        dp = [-1] * n
        @lru_cache(None)
        def dfs(i, c):
            if i > L:
                return c == 0
            for d in range(9, 0, -1):
                c1 = (d * m[i]) % k
                if not(i == L and n % 2 == 1):
                    c1 += (d * m[n - 1 - i]) % k
                c1 = (c + c1) % k
                # print(c, c1)
                res = dfs(i + 1, c1)
                if res:
                    dp[i] = d
                    return True
            return False
        
        arr = dfs(0, 0)
        for i in range(n - 1, -1, -1):
            if dp[i] != -1:
                break
            dp[i] = dp[n - 1 - i]
        return ''.join(str(c) for c in dp)


            


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