<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:
    L = 10;

    def getDigit(self, nucleotide):
        if nucleotide == 'A':
            return 0;
        elif nucleotide == 'C':
            return 1;
        elif nucleotide == 'G':
            return 2;
        else:
            return 3;

    def findRepeatedDnaSequences(self, s: str) -> List[str]:
        seen = set();
        left, right = 0, 0;
        duplicates = set();
        res = set();
        window = 0;
        R = 4;
        L = 10;
        RL = pow(R,(L-1));

        while right < len(s):
            c = self.getDigit(s[right]);
            window = window * R + c;
            print("window", window, window * RL)
            right += 1;

            if right - left == self.L:
                d = self.getDigit(s[left]);
                if window in seen:
                    res.add(s[left:right]);
                seen.add(window);
                window = window - RL*d;
                left += 1;

        return list(res);

        
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