<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]:
        if len(s) <= 10: return []
        
        seen = set()
        ans = set()

        d = {'A' : 0, 'C' : 1, 'T' : 2, 'G' : 3}
        d2 = {0 : 'A', 1: 'C', 2: 'T', 3: 'G'}

        def decode(num: int):
            ans = ''
            for i in range(10):
                ans = d2[num & 3] + ans
                num = num >> 2
            return ans

        curr = 0
        for i in range(10): curr = curr << 2 | d[s[i]]

        seen.add(curr)
        for i in range(10, len(s)):
            curr = (curr << 2 | d[s[i]]) & ((1 << 20) - 1)
            if curr in seen: ans.add(curr)
            seen.add(curr)

        return [decode(a) for a in ans]
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