<p>Given a string <code>s</code>, consider all <em>duplicated substrings</em>: (contiguous) substrings of s that occur 2 or more times.&nbsp;The occurrences&nbsp;may overlap.</p>

<p>Return <strong>any</strong> duplicated&nbsp;substring that has the longest possible length.&nbsp;If <code>s</code> does not have a duplicated substring, the answer is <code>&quot;&quot;</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> s = "banana"
<strong>Output:</strong> "ana"
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> s = "abcd"
<strong>Output:</strong> ""
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= s.length &lt;= 3 * 10<sup>4</sup></code></li>
	<li><code>s</code> consists of lowercase English letters.</li>
</ul>

class Solution:
    def RabinKarp(self,text, M, q):
        if M == 0: return True
        h, t, d = (1<<(8*M-8))%q, 0, 256

        dic = defaultdict(list)

        for i in range(M): 
            t = (d * t + ord(text[i]))% q

        dic[t].append(i-M+1)

        for i in range(len(text) - M):
            t = (d*(t-ord(text[i])*h) + ord(text[i + M]))% q
            for j in dic[t]:
                if text[i+1:i+M+1] == text[j:j+M]:
                    return (True, text[j:j+M])
            dic[t].append(i+1)
        return (False, "")

    def longestDupSubstring(self, S):
        beg, end = 0, len(S)
        q = (1<<31) - 1 
        Found = ""
        while beg + 1 < end:
            mid = (beg + end)//2
            isFound, candidate = self.RabinKarp(S, mid, q)
            if isFound:
                beg, Found = mid, candidate
            else:
                end = mid

        return Found

        
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