A DNA is composed of a series of nucleotides abbreviated as A, C, G, and T, for example: 'ACGAATTCCG'. Write a function to find all the 10-letter-long sequences(sub-strings) that occur more than once in a DNA molecule. Example - Input: 'AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT', Expected Output: [AAAAACCCCC,CCCCCAAAAA]

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Algorithm/Insights

A naive approach of sliding the 10-letter window across the given sequence combined with hashmap would take O(n^2) time. Using rolling hash method takes O(n) time. Here are the steps. 1. Compute hash value for the sequence in the first window. 2. Store the computed value in a set/hashmap. 3. Compute hash values for subsequent sequence(which would be sequence obtained by sliding 10-letter window to the right by one character) using rolling hash method. Rolling hash method makes use of hash value computed for previous sequence to compute hash value for current sequence. 4. If computed hash value is already present in the hashmap then add the current sequence to output set, else store the computed value in the hashmap. . 5. Repeat step #3, #4 until all the 10-letter sequences are completed.

Rolling Hash computation uses following method. For details and intuition, please check out the video. currHash = prevHash - val(skippedChar)*2^10 currHash = currHash * 2 currHash = currHash + 2*val(newChar)

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Algorithm Visualization

Code Snippet

package com.ideserve.nilesh.questions;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.Set;
public class RepeatedDnaSequencesSolution
{
private static final Map<Character, Integer> encodings = new HashMap<>();
static { encodings.put('A',0); encodings.put('C',1); encodings.put('G',2); encodings.put('T',3); }
private final int Two_POW_9 = (int) Math.pow(2, 9);
public List<String> findRepeatedDnaSequences(String s) {
Set<String> res = new HashSet<>();
Set<Integer> hashes = new HashSet<>();
List<String> solution = new ArrayList<String>();
HashMap duplicates = new HashMap<Integer, String>();
for (int i = 0, rhash = 0; i < s.length(); i++)
{
if (i > 9) rhash -= Two_POW_9 * encodings.get(s.charAt(i-10));
rhash = 2 * rhash + encodings.get(s.charAt(i));
if (i > 8)
{
if (duplicates.get(rhash) != null)
{
res.add(s.substring(i-9,i+1));
}
else
{
duplicates.put(rhash, "");
}
}
}
return new ArrayList<>(res);
}
public static void main(String[] args)
{
RepeatedDnaSequencesSolution sln = new RepeatedDnaSequencesSolution();
List<String> list = sln.findRepeatedDnaSequences("AACAAAAACAAAACCAAAAACAAAAACAAAA");
System.out.println("Repeated DNA sequences are: ");
for (int i = 0; i < list.size(); i++)
{
System.out.println(list.get(i));
}
}
}

Order of the Algorithm

Time Complexity is O(n) Space Complexity is O(n)

Contribution

Sincere thanks from IDeserve community to Nilesh More for compiling current post.

Nilesh More

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