Shortest Palindrome
<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>
String; String Matching; Hash Function; Rolling Hash
class Solution:
    def shortestPalindrome(self, s: str) -> str:
        base = 131
        mod = 10**9 + 7
        n = len(s)
        prefix = suffix = 0
        mul = 1
        idx = 0
        for i, c in enumerate(s):
            prefix = (prefix * base + (ord(c) - ord('a') + 1)) % mod
            suffix = (suffix + (ord(c) - ord('a') + 1) * mul) % mod
            mul = (mul * base) % mod
            if prefix == suffix:
                idx = i + 1
        return s if idx == n else s[idx:][::-1] + s
