<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 n = s.size();
        vector<int> fail(n, -1);
        for (int i = 1; i < n; ++i) {
            int j = fail[i - 1];
            while (j != -1 && s[j + 1] != s[i]) {
                j = fail[j];
            }
            if (s[j + 1] == s[i]) {
                fail[i] = j + 1;
            }
        }
        int best = -1;
        for (int i = n - 1; i >= 0; --i) {
            while (best != -1 && s[best + 1] != s[i]) {
                best = fail[best];
            }
            if (s[best + 1] == s[i]) {
                ++best;
            }
        }
        string add = (best == n - 1 ? "" : s.substr(best + 1, n));
        reverse(add.begin(), add.end());
        return add + s;
    }
};
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