<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>

int solve(string &s){
    int n = s.size() ;
    int i = 0 ;
    vector<int> dp(n) ;
    for(int e=1 ;e<n ;e++){
        if(s[i]==s[e]){
            dp[e] = dp[e-1]+1 ;
            i++ ; 
        }
        else{
            int j ;
            l:;
            while(i>=0 && s[e]!=s[i])i-- ;
            if(i<0)i++ ;
            else{
                j= e ;
                while(i>=0 && s[i]==s[j])i-- , j-- ;
                if(i<0){
                    i = (e-j) ;
                    dp[e]= i ;
                }
                else{
                    goto l;
                }
            }
        }
    }
    int ans = n-1 ;
    // for(int e=0 ;e<n ;e++){
    //     if(dp[ans]<dp[e])ans = e ;
    // }
    return dp[ans] ;
}
class Solution {
public:
    string longestPrefix(string s) {
        int len=solve(s) ;
        while(s.size()>len)s.pop_back() ;
        return s;
    }
};
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