<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) {
        long long frontHash = 0, backHash = 0, p = 31, mod = 1e9 + 9, power = 1;
        int ind = -1;
        int n = s.length();
        int i = 0, j = n - 1;
        while(i < n-1 && j > 0){
            frontHash = (frontHash + (s[i] * power) % mod) % mod;
            power = (power * p) % mod;

            backHash = (backHash * p) % mod;
            backHash = (backHash + s[j]) % mod;

            if(frontHash == backHash){
                ind = i;
            }
            i++;
            j--;
        }

        string ans = "";
        for(int x = 0; x <= ind; x++){
            ans.push_back(s[x]);
        }
        return ans;
    }
};
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