<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 {
    int hashVal(char c) {
        switch (c) {
        case 'A':
            return 0;
        case 'C':
            return 1;
        case 'G':
            return 1;
        case 'T':
            return 1;
        default:
            return 0;
        }
    }

public:
    vector<string> findRepeatedDnaSequences(string s) {
        if (s.length() < 10)
            return {};

        unordered_map<int, vector<int>> hash_map;
        unordered_set<string> result_set;

        int hash = 0;
        int base = 4;
        int window = 10;

        int highestbase = pow(base, window - 1);
        for (int i = 0; i < window; i++) {
            hash = hash * base + hashVal(s[i]);
        }
        hash_map[hash].push_back(0);

        for (int i = window; i < s.length(); i++) {
            hash = (hash - hashVal(s[i - window]) * highestbase) * base +
                   hashVal(s[i]);
            if (hash_map.find(hash) != hash_map.end()) {

                for (int start_index : hash_map[hash]) {
                    if (s.substr(start_index, window) ==
                        s.substr(i - window + 1, window)) {
                        result_set.insert(s.substr(i - window + 1, window));

                        break;
                    }
                }
            }
            hash_map[hash].push_back(i - window + 1);
        }

        vector<string> result(result_set.begin(), result_set.end());

        return result;
    }
};
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