<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 {
public:

    /*inline int DNAInt(char l)
    {
        switch (l)
        {
        case 'A':
            return 0;
        case 'C':
            return 1;
        case 'G':
            return 2;
        case 'T':
        default:
            return 3;
        }
    }*/

    struct HashTrack
    {
        bool seen = false;
        bool repeat = false;
    };

    vector<string> findRepeatedDnaSequences(string s) {
        constexpr int NUMHASHES = 4 * 4 * 4 * 4 * 4 * 4 * 4 * 4 * 4 * 4;

        int n = s.length();
        if (n <= 10)
            return vector<string>();
        
        HashTrack hashes[NUMHASHES] {};
        int DNAInt[85];
        DNAInt['A'] = 0;
        DNAInt['C'] = 1;
        DNAInt['G'] = 2;
        DNAInt['T'] = 3;

        int hash = 0;
        for (int i = 0; i < 10; ++i)
        {
            hash <<= 2;
            hash += DNAInt[s[i]];
        }
        hashes[hash].seen = true;

        vector<string> result;
        for (int i = 10; i < n; ++i)
        {
            hash &= 0b00111111111111111111;
            hash <<= 2;
            hash += DNAInt[s[i]];
            if (hashes[hash].seen && !hashes[hash].repeat)
            {
               result.emplace_back(s.substr(i - 9, 10));
               hashes[hash].repeat = true;
            }
            else
               hashes[hash].seen = true;
        }

        return result;
    }
};
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