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

        int j = 0, endi = 0; 
        i = s.size()-1;

        while(i>=0){
            if(s[i] == s[j]){
                i--;j++;
            }
            else if(j == 0) i--;
            else j = lps[j-1];
            if(i == j){
                endi = 2*i;
                break;
            }
            else if(i<j){
                endi = 2*i + 1; 
                break;
            }
        }
        endi++;
        string p = s.substr(endi,s.size() - endi +1);
        reverse(p.begin(), p.end());
        p = p + s;
        return p;
    }
};
10300