Find the Closest Palindrome
<p>Given a string <code>n</code> representing an integer, return <em>the closest integer (not including itself), which is a palindrome</em>. If there is a tie, return <em><strong>the smaller one</strong></em>.</p>

<p>The closest is defined as the absolute difference minimized between two integers.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = &quot;123&quot;
<strong>Output:</strong> &quot;121&quot;
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = &quot;1&quot;
<strong>Output:</strong> &quot;0&quot;
<strong>Explanation:</strong> 0 and 2 are the closest palindromes but we return the smallest which is 0.
</pre>

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

<ul>
	<li><code>1 &lt;= n.length &lt;= 18</code></li>
	<li><code>n</code> consists of only digits.</li>
	<li><code>n</code> does not have leading zeros.</li>
	<li><code>n</code> is representing an integer in the range <code>[1, 10<sup>18</sup> - 1]</code>.</li>
</ul>
Math; String
class Solution:
    def nearestPalindromic(self, n: str) -> str:
        x = int(n)
        l = len(n)
        res = {10 ** (l - 1) - 1, 10**l + 1}
        left = int(n[: (l + 1) >> 1])
        for i in range(left - 1, left + 2):
            j = i if l % 2 == 0 else i // 10
            while j:
                i = i * 10 + j % 10
                j //= 10
            res.add(i)
        res.discard(x)

        ans = -1
        for t in res:
            if (
                ans == -1
                or abs(t - x) < abs(ans - x)
                or (abs(t - x) == abs(ans - x) and t < ans)
            ):
                ans = t
        return str(ans)
