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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0101_0200.s0187_repeated_dna_sequences;

// #Medium #String #Hash_Table #Bit_Manipulation #Sliding_Window #Hash_Function #Rolling_Hash
// #Data_Structure_II_Day_9_String #Udemy_Strings
// #2022_06_27_Time_29_ms_(77.11%)_Space_74.1_MB_(6.94%)

import java.util.ArrayList;
import java.util.Collections;
import java.util.List;

/**
 * 187 - Repeated DNA Sequences\.
 *
 * Medium
 *
 * The **DNA sequence** is composed of a series of nucleotides abbreviated as `'A'`, `'C'`, `'G'`, and `'T'`.
 *
 * *   For example, `"ACGAATTCCG"` is a **DNA sequence**.
 *
 * When studying **DNA** , it is useful to identify repeated sequences within the DNA.
 *
 * Given a string `s` that represents a **DNA sequence** , return all the **`10`\-letter-long** sequences (substrings) that occur more than once in a DNA molecule. You may return the answer in **any order**.
 *
 * **Example 1:**
 *
 * **Input:** s = "AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT"
 *
 * **Output:** ["AAAAACCCCC","CCCCCAAAAA"] 
 *
 * **Example 2:**
 *
 * **Input:** s = "AAAAAAAAAAAAA"
 *
 * **Output:** ["AAAAAAAAAA"] 
 *
 * **Constraints:**
 *
 * *   1 <= s.length <= 10⁵
 * *   `s[i]` is either `'A'`, `'C'`, `'G'`, or `'T'`.
**/
public class Solution {
    public List findRepeatedDnaSequences(String s) {
        if (s.length() < 10) {
            return Collections.emptyList();
        }
        boolean[] seen = new boolean[1024 * 1024];
        boolean[] added = new boolean[1024 * 1024];
        char[] chars = s.toCharArray();
        int buf = 0;
        int[] map = new int[128];
        map['A'] = 0;
        map['C'] = 1;
        map['G'] = 2;
        map['T'] = 3;
        List ans = new ArrayList<>(s.length() / 2);
        for (int i = 0; i < 10; i++) {
            buf = (buf << 2) + map[chars[i]];
        }
        seen[buf] = true;
        for (int i = 10; i < chars.length; i++) {
            buf = ((buf << 2) & 0xFFFFF) + map[chars[i]];
            if (seen[buf]) {
                if (!added[buf]) {
                    ans.add(new String(chars, i - 9, 10));
                    added[buf] = true;
                }
            } else {
                seen[buf] = true;
            }
        }
        return ans;
    }
}