<p>You are given a string <code>s</code>. You can convert <code>s</code> to a <span data-keyword="palindrome-string">palindrome</span> by adding characters in front of it.</p>

<p>Return <em>the shortest palindrome you can find by performing this transformation</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> s = "aacecaaa"
<strong>Output:</strong> "aaacecaaa"
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> s = "abcd"
<strong>Output:</strong> "dcbabcd"
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>0 &lt;= s.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>s</code> consists of lowercase English letters only.</li>
</ul>

class Solution:
    def shortestPalindrome(self, s: str) -> str:
        if len(s) in [0, 1]:
            return s

        left = ord(s[0]) - ord("a")
        left_mul = 26

        right = ord(s[1]) - ord("a")
        right_mul = 1
        mid = 1

        longest_palindrome = 1 if left != right else 2

        for i, ch in enumerate(s[2:], 2):
            if i % 2 == 0:
                right = right - (ord(s[mid]) - ord("a")) * right_mul
                right = right * 26 + ord(ch) - ord("a")
                right_mul *= 26
            else:
                left += (ord(s[mid]) - ord("a")) * left_mul
                mid += 1
                left_mul *= 26

                right = 26 * right + ord(ch) - ord("a")
            
            if left == right:
                longest_palindrome = i + 1

        return s[longest_palindrome:][::-1] + s

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