<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:
    vector<string> findRepeatedDnaSequences(string s) {
        
        int L = 10;
        int str_length = s.length();
        vector<int> string_number;

        if(str_length < L) return vector<string>();

        int a = 4;
        int ha = pow(a, L);

        //convert the given string
        map<char, int> char_to_int {{'A', 0}, {'C', 1}, {'T', 2}, {'G', 3}};

        unordered_set<int> seen;
        unordered_set<string> output;
        int h = 0;

        for(auto& c : s)
        {
            string_number.push_back(char_to_int[c]);
        }

        //now itr over every sub string of length 10 
        for(int i = 0; i < str_length - L + 1; i++)
        {
            if(i == 0)
            {
                //the first computation
                for(int d = 0; d < L; d++)
                {
                    h = h * a + string_number[d];
                }

            }else{

                //compute based on prev computation 
                h = h * a - string_number[i - 1] * ha + string_number[i + L - 1];
            }

            if(seen.find(h) != seen.end()) output.insert(s.substr(i, L));
            seen.insert(h);
        }

        return vector<string>(output.begin(), output.end());

    }
};
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