<p>The <strong>DNA sequence</strong> is composed of a series of nucleotides abbreviated as <code>&#39;A&#39;</code>, <code>&#39;C&#39;</code>, <code>&#39;G&#39;</code>, and <code>&#39;T&#39;</code>.</p>

<ul>
	<li>For example, <code>&quot;ACGAATTCCG&quot;</code> is a <strong>DNA sequence</strong>.</li>
</ul>

<p>When studying <strong>DNA</strong>, it is useful to identify repeated sequences within the DNA.</p>

<p>Given a string <code>s</code> that represents a <strong>DNA sequence</strong>, return all the <strong><code>10</code>-letter-long</strong> sequences (substrings) that occur more than once in a DNA molecule. You may return the answer in <strong>any order</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> s = "AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT"
<strong>Output:</strong> ["AAAAACCCCC","CCCCCAAAAA"]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> s = "AAAAAAAAAAAAA"
<strong>Output:</strong> ["AAAAAAAAAA"]
</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[i]</code> is either <code>&#39;A&#39;</code>, <code>&#39;C&#39;</code>, <code>&#39;G&#39;</code>, or <code>&#39;T&#39;</code>.</li>
</ul>

class Solution:
    def findRepeatedDnaSequences(self, s: str) -> List[str]:
        base=4
        mapping={'A':1, 'C':2, 'G':3, 'T':4}
        max_coeff=pow(base,9)
        nums=[]
        n=len(s)
        for i in range(0,n,1):
            nums.append(mapping[s[i]])
        seen=set()
        ans=set()
        hashing=0
        start=0
        while start<n-9:
            if start!=0:
                hashing-=(nums[start-1]*max_coeff)
                hashing*=base
                hashing+=nums[start+9]
            else:
                for i in range(0,10,1):
                    hashing=hashing*base+nums[i]
            if hashing in seen:
                ans.add(s[start:start+10])
            seen.add(hashing)
            start+=1
        return ans
26800