<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:

void help(string s,int lps[]){
      lps[0]=0;
      int len=0;
      int n=s.length();

      for(int i=1;i<n;){
          if(s[i]==s[len]){
              len++;
              lps[i]=len;
              i++;
          }
          else {
                  if(len>0)
                  len=lps[len-1];
              else{
                   lps[i]=0;
                   i++;
                   len=0;
              }
                 
              }
          }
      
      return;
  }
    string longestPrefix(string s) {
        int n=s.length();
        int lps[n];
        help(s,lps);
        int max_len=lps[n-1],ind=n-1;
        // for(int i=1;i<n;i++){
        //     if(lps[i]>max_len){
        //         max_len=lps[i];
        //         ind=i;
        //     }
        // }
        string ans="";
        int s_ind=(ind-max_len)+1;
        while(max_len){
            ans+=s[s_ind];
            s_ind++;
            max_len--;
        }
return ans;
    }
};
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