<p>A string is called a <strong>happy prefix</strong> if is a <strong>non-empty</strong> prefix which is also a suffix (excluding itself).</p>

<p>Given a string <code>s</code>, return <em>the <strong>longest happy prefix</strong> of</em> <code>s</code>. Return an empty string <code>&quot;&quot;</code> if no such prefix exists.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;level&quot;
<strong>Output:</strong> &quot;l&quot;
<strong>Explanation:</strong> s contains 4 prefix excluding itself (&quot;l&quot;, &quot;le&quot;, &quot;lev&quot;, &quot;leve&quot;), and suffix (&quot;l&quot;, &quot;el&quot;, &quot;vel&quot;, &quot;evel&quot;). The largest prefix which is also suffix is given by &quot;l&quot;.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;ababab&quot;
<strong>Output:</strong> &quot;abab&quot;
<strong>Explanation:</strong> &quot;abab&quot; is the largest prefix which is also suffix. They can overlap in the original string.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>s</code> contains only lowercase English letters.</li>
</ul>

class Solution {
public:
    int compute_pi(int i, vector <int> &pi, string &P){
    if(i<=1) return 0;
    //pi[i] -> length of the longest prefix of P[1...i], which is also a suffix of P[1...i]
    int k = pi[i-1];
    while(P[k+1]!=P[i] && k>0) k = pi[k];
    if(P[k+1]==P[i]) return k+1;
    return 0;
    }
    string longestPrefix(string s) {
        string temp;
        temp.push_back('a');
        for(int i=0; i<s.size(); i++) temp.push_back(s[i]);
        int m = s.size();
        vector <int> pi(m+1, 0);
        for(int i=1; i<=m; i++) pi[i] = compute_pi(i, pi, temp);
        string ans;
        for(int i=1; i<=pi[m]; i++) ans.push_back(temp[i]);
        return ans;
    }
};
24990