<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 size = s.size();

        std::cout << "kmp: ";
        /* KMP-Table Building Algorithm */
        vector<int> kmp_table(size, 0);
        //std::cout << kmp_table[0] << ",";
        for (int i = 1; i < size; ++i) {
            int j = kmp_table[i-1];
            while (j > 0 && s[i] != s[j]) {
                j = kmp_table[j-1];
            }
            if (s[i] == s[j]) {
                j++;
            }
            kmp_table[i] = j;
            //std::cout << kmp_table[i] << ",";
        }
        //std::cout << std::endl << "f: ";

        /* Now that we have it, we then reverse string and find largest match */

        std::string rev(s);
        reverse(rev.begin(), rev.end());


        vector<int> f(rev.size(), 0);
        //std::cout << f[0] << ",";
        for (int i = 0; i < rev.size(); i++) {
            int t = 0;
            if (i != 0) t = f[i-1];
            while (t > 0 && rev[i] != s[t]) {
                t = kmp_table[t-1];
            }
            if (rev[i] == s[t]) {
                ++t;
            }
            f[i] = t;
            //std::cout << f[i] << ",";
        }

        // Return reverse of matching string + s
        if (rev.size() == 0) return s;
        return rev.substr(0, size - f[rev.size()-1]) + s;
    }
};
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