<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 {
    vector<int> lpsTable(string p) {
        int n = p.size();
        vector<int> lps(n);
        int i = 1, idx = 0;
        while(i<n) {
            if(p[i] == p[idx]) {
                lps[i] = idx+1;
                i++, idx++;
            } else if(idx>0) {
                idx = lps[idx-1];
            } else {
                lps[i] = 0;
                i++;
            }
        }
        return lps;
    }

public:
    string shortestPalindrome(string s) {
        string rev = s;
        reverse(rev.begin(), rev.end());
        vector<int> lps = lpsTable(s);

        int i = 0, j = 0;
        while(i<rev.size()) {
            if(rev[i] == s[j]) {
                i++, j++;
            } else if(j > 0) {
                j = lps[j-1];
            } else {
                i++;
            }
        }

        string tmp = s.substr(j);
        reverse(tmp.begin(), tmp.end());
        return tmp + s;
    }
};
10600