<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 {
public:
    string shortestPalindrome(string s) {
        int length = s.length();
        if (length == 0) {
            return s;
        }

        // Find the longest palindromic prefix
        int left = 0;
        for (int right = length - 1; right >= 0; right--) {
            if (s[right] == s[left]) {
                left++;
            }
        }

        // If the whole string is a palindrome, return the original string
        if (left == length) {
            return s;
        }

        // Extract the suffix that is not part of the palindromic prefix
        string nonPalindromeSuffix = s.substr(left);
        string reverseSuffix =
            string(nonPalindromeSuffix.rbegin(), nonPalindromeSuffix.rend());

        // Form the shortest palindrome by prepending the reversed suffix
        return reverseSuffix + shortestPalindrome(s.substr(0, left)) +
               nonPalindromeSuffix;
    }
};
10000