<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) {
        string rev = s;
        reverse(rev.begin(), rev.end());
        auto lps = getNext(s);
        int j = 0;
        for(int i = 0; i < rev.size(); ++i){
            while(j > 0 && rev[i] != s[j])
                j = lps[j - 1];
            if(rev[i] == s[j])
                ++j;
        }
        return rev + s.substr(j);
    }
    vector<int> getNext(const string& s){
        vector<int> res(s.size());
        for(int i = 1, j = 0; i < s.size(); ++i){
            while(j > 0 && s[i] != s[j])
                j = res[j - 1];
            if(s[i] == s[j])
                ++j;
            res[i] = j;
        }
        return res;
    }
};

/*

rev: aaacecaa
s:    aacecaaa

*/
10400