<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 s == s[::-1]:
            return s 
        possible_pivots = [i for i in range(len(s)//2)][::-1]
        print(possible_pivots)
        for pivot in possible_pivots:
            if s[pivot:pivot+2] == s[pivot:pivot+2][::-1]: #basically if there is a dual pivot 
                if s[:pivot] == s[pivot+2:pivot+2+len(s[:pivot])][::-1]:
                    return s[pivot+2+len(s[:pivot])::][::-1] + s

            if s[:pivot] == s[pivot+1:pivot+1+len(s[:pivot])][::-1]:
                return s[pivot+1+len(s[:pivot])::][::-1] + s 
17800