<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 create_LPS(string s, int LPS[]) {
        int i = 1, idx = 0, n = s.size();
        while (i < n) {
            if (s[i] == s[idx]) {
                LPS[i] = idx + 1;
                i++;
                idx++;
            } else {
                if (idx > 0) {
                    idx = LPS[idx - 1];
                } else {
                    i++;
                }
            }
        }
    }

    string longestPrefix(string s) {
        ios_base::sync_with_stdio(false);
        cin.tie(0);

        string str = s;
        int n = str.size();

        int LPS[n];
        memset(LPS, 0, sizeof(LPS));

        create_LPS(str, LPS);

        // for (auto& it : str) {
        //     cout << it << " ";
        // }
        // cout << "\n";

        // for (auto& it : LPS) {
        //     cout << it << " ";
        // }
        // cout << "\n";

        return s.substr(0, LPS[n - 1]);
    }
};
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