<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>

from collections import deque
class Solution:
    def findRepeatedDnaSequences(self, s: str) -> List[str]:
        
        a=list(s)
        # print(s,len(s),len(a))
        # s = "AAAAAAAAAAAAA"
        n=len(s)
        i=j=c=1
        l=deque()
        k=10
        s1=a[:10]
        s1=deque(s1)
        s2=''.join(s1)
        # print(s2)
        if(s2 in s[1:]):
            l.append(s2)

        # print(s1)
        i=0
        while(i<n-10):
            # if(i<n):
            # print(i,s1,s1[i])
            # del s1[0]
            s1.popleft()
            # print(s1)
            s1.append(s[i+10])
            s2=''.join(s1)
            # print(s2)
            if(s2 in s[j+1:]):
                l.append(s2)
            i+=1
            j+=1
        return set(set(l))
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