<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>

// KMP
class Solution {
public:
    string shortestPalindrome(string p) {
        string newString = p;
        string rs = p;
        reverse(rs.begin(), rs.end());
        newString += '#';
        newString += rs;

        int sz = newString.size();
        int pi[sz];
        pi[0] = 0;
        for(int i = 1, k = 0; i < sz; i++) {
            while(k > 0 && newString[i] != newString[k]) k = pi[k-1];
            if(newString[i] == newString[k]) k++;
            pi[i] = k;
        }
        cout << newString << endl;
        for(int i = 0; i < sz; i++) cout << pi[i] << " ";
        string res = rs.substr(0, rs.size() - pi[sz-1]);
        res += p;
        return res;
    }
};

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