<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:
        # Reverse the original string
        reversed_string = s[::-1]

        # Combine the original and reversed strings with a separator
        combined_string = s + "#" + reversed_string

        # Build the prefix table for the combined string
        prefix_table = self._build_prefix_table(combined_string)

        # Get the length of the longest palindromic prefix
        palindrome_length = prefix_table[-1]

        # Construct the shortest palindrome by appending the reverse of the suffix
        suffix = reversed_string[: len(s) - palindrome_length]
        return suffix + s

    # Helper function to build the KMP prefix table
    def _build_prefix_table(self, s: str) -> list:
        prefix_table = [0] * len(s)
        length = 0

        # Build the table by comparing characters
        for i in range(1, len(s)):
            while length > 0 and s[i] != s[length]:
                length = prefix_table[length - 1]
            if s[i] == s[length]:
                length += 1
            prefix_table[i] = length
        return prefix_table
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