<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:
    string longestPrefix(string s) {
        //dp[i] = len of the longest proper prefix which 
        //is also proper suffix in the string of len 0 till i

        //proper means excluding the string itself 

         int len = 0;
         int idx = 1;
         vector<int>dp(s.length(),0);
         dp[0] = len;
         while(idx < s.length())
         {
              if(s[len] == s[idx]){
                 len++;
                 dp[idx] = len;
                 idx++;
              }   
              else if(s[len] != s[idx]){
                  if(len-1 >= 0){
                      len = dp[len-1];
                  }
                  else{
                      dp[idx] = 0;
                      idx++;
                  }
              }
         }
         if(dp.back() == 0) return "";
         string ans = "";
         int properPrefixLen = dp.back();
         for(int i = 0 ; i < s.length() and properPrefixLen > 0 ; i++){
             ans.push_back(s[i]);
             properPrefixLen--;
         }
         return ans;
        
    }
};
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