Longest Happy Prefix
<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>
String; String Matching; Hash Function; Rolling Hash
class Solution {
    private long[] p;
    private long[] h;

    public String longestPrefix(String s) {
        int base = 131;
        int n = s.length();
        p = new long[n + 10];
        h = new long[n + 10];
        p[0] = 1;
        for (int i = 0; i < n; ++i) {
            p[i + 1] = p[i] * base;
            h[i + 1] = h[i] * base + s.charAt(i);
        }
        for (int l = n - 1; l > 0; --l) {
            if (get(1, l) == get(n - l + 1, n)) {
                return s.substring(0, l);
            }
        }
        return "";
    }

    private long get(int l, int r) {
        return h[r] - h[l - 1] * p[r - l + 1];
    }
}