<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 MOD=1e9+7;
    string longestPrefix(string s) {
        int n=s.size();
        long long a=0;
        long long b=0;
        vector<long long> pref(n,0);
        long long k=1;
        
        for(int i=0;i<n-1;i++){
            a=((31*a)%MOD+(s[i]-'a'+1));
           
            pref[i]=a;
            
        }
        
        int ans=-1;
        
        for(int i=n-1;i>=1;i--){
            b=(b+(k*(s[i]-'a'+1))%MOD)%MOD;
            
            if(pref[n-i-1]==b){
                ans=n-i-1;
            }
            k=(k*31)%MOD;
        }
        return ans!=-1?s.substr(0,ans+1):"";
    }
};
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