<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]:
        hmap = {'A': 0, 'G': 1, 'C': 2, 'T': 3}

        if len(s) < 10:
            return []

        hash1 = 0

        for i in range(len(s)):
            hash1 += hmap[s[i]]

        rst = {}

        rst[hash1] = [s[0:10]]

        res = set()


        for i in range(1, len(s) - 10 + 1):
            hash1 -= hmap[s[i - 1]]
            hash1 += hmap[s[i + 9]]
            if hash1 in rst and s[i: i + 10] in rst[hash1]:
                res.add(s[i: i + 10])
            else:
                if hash1 not in rst:
                    rst[hash1] = []
                rst[hash1].append(s[i: i + 10])

        return list(res)
            



        
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