<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:
        # Step 1: Create a new string "s + # + reverse(s)"
        rev_s = s[::-1]
        new_s = s + '#' + rev_s
        
        # Step 2: Compute the KMP table (prefix function)
        lps = [0] * len(new_s)  # lps array for storing the longest prefix suffix lengths
        for i in range(1, len(new_s)):
            j = lps[i - 1]  # length of the previous longest prefix suffix
            while j > 0 and new_s[i] != new_s[j]:
                j = lps[j - 1]
            if new_s[i] == new_s[j]:
                j += 1
            lps[i] = j
        
        # Step 3: Get the length of the longest palindromic prefix
        longest_palindromic_prefix_len = lps[-1]
        
        # Step 4: Append the reverse of the remaining suffix in front
        remaining_suffix = s[longest_palindromic_prefix_len:]
        return remaining_suffix[::-1] + s

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