Gene Mutation Groups - Google Top Interview Questions

Problem Statement :

You are given a list of unique strings genes where each element has the same length and contains characters "A", "C", "G" and/or "T".

If strings a and b are the same string except for one character, then a and b are in the same mutation group.

If strings a and b are in a group and b and c are in a group, then a and c are in the same group.
Return the total number of mutation groups.


n ≤ 10,000

k ≤ 20 where k is the length of a string in genes

Example 1


genes = ["ACGT", "ACCT", "AGGT", "TTTT", "TTTG"]




There are two mutation groups:

["ACGT", "ACCT", "AGGT"]

["TTTT", "TTTG"]

Solution :


                        Solution in C++ :

class UnionFind {
    vector<int> parents, rank;

    UnionFind(int n) {
        for (int i = 0; i < n; i++) {
            parents[i] = i;
            rank[i] = 1;

    int find(int node) {
        int root = node;

        while (root != parents[root]) {
            root = parents[root];

        // Path compression
        while (node != root) {
            int temp = parents[node];
            parents[node] = root;
            node = temp;

        return root;

    // Returns true if union happens
    bool unify(int a, int b) {
        int rootA = find(a);
        int rootB = find(b);

        if (rootA == rootB) return false;

        // Union by rank
        if (rank[rootA] > rank[rootB]) {
            parents[rootB] = rootA;
        } else if (rank[rootB] > rank[rootA]) {
            parents[rootA] = rootB;
        } else {
            parents[rootB] = rootA;

        return true;

int solve(vector<string>& genes) {  // Time: O(N * K), Space: O(N)
    int n = genes.size();
    unordered_map<string, int> gene_map;
    char type[4] = {'A', 'C', 'G', 'T'};

    for (int i = 0; i < n; i++) gene_map[genes[i]] = i;

    UnionFind union_find(n);
    int groups = n;

    for (string& gene : genes) {
        int curr = gene_map[gene];

        for (char& c : gene) {
            char orig = c;
            for (char t : type) {
                if (t != orig) {
                    c = t;
                    if (gene_map.count(gene)) {
                        if (union_find.unify(curr, gene_map[gene])) groups--;

            c = orig;

    return groups;

                        Solution in Java :

import java.util.*;

class Solution {
    class DisjointSet {
        int val;
        DisjointSet parent;
        public DisjointSet(int value) {
            this.val = value;
            this.parent = this;
    private static final String delim = "_";
    private Map<Integer, DisjointSet> map = new HashMap();
    private Map<String, Integer> indexMap = new HashMap();
    private Set<DisjointSet> parent = new HashSet();

    private String s1 = "CGT";
    private String s2 = "AGT";
    private String s3 = "ACT";
    private String s4 = "ACG";
    private Map<Character, String> replaceMap = new HashMap();

    public int solve(String[] genes) {
        if (genes == null || genes.length == 0)
            return 0;

        replaceMap.put('A', s1);
        replaceMap.put('C', s2);
        replaceMap.put('G', s3);
        replaceMap.put('T', s4);

        for (int i = 0; i < genes.length; i++) map.put(i, new DisjointSet(i));

        for (int i = 0; i < genes.length; i++) {
            String str = genes[i];
            for (int j = 0; j < str.length(); j++) {
                String replaceString = replaceMap.get(str.charAt(j));
                for (int k = 0; k < replaceString.length(); k++) {
                    char replaceChar = replaceString.charAt(k);
                    String temStr = new StringBuilder().append(str).toString();
                    char[] tempChar = temStr.toCharArray();
                    tempChar[j] = replaceChar;
                    if (indexMap.containsKey(new String(tempChar))) {
                        union(i, indexMap.get(new String(tempChar)));
            indexMap.put(str, i);

        for (int i = 0; i < genes.length; i++) {
            // add the parent (i.e find) of each node in a hashset
        return parent.size();

    private void union(int idx1, int idx2) {
        DisjointSet set1 = map.get(idx1);
        DisjointSet set2 = map.get(idx2);

        DisjointSet par1 = find(set1);
        DisjointSet par2 = find(set2);

        if (par1.val == par2.val)
        par1.parent = par2;

    private DisjointSet find(DisjointSet set) {
        if (set.parent == set)
            return set;
        return set.parent = find(set.parent);

                        Solution in Python : 
class Solution:
    def solve(self, A):
        dsu = DSU()
        for word in A:
            for i in range(len(word)):
                root = word[:i] + "*" + word[i + 1 :]
                dsu.union(word, root)
        return len(set(map(dsu.find, A)))

class DSU:
    def __init__(self): = {}
        self.par = [] = []

    def find(self, x):
            i =[x]
  [x] = i = len(
        return self._find(i)

    def _find(self, x):
        if self.par[x] != x:
            self.par[x] = self._find(self.par[x])
        return self.par[x]

    def union(self, x, y):
        xr, yr = self.find(x), self.find(y)
        if xr == yr:
            return False
        if[xr] <[yr]:
            xr, yr = yr, xr
        self.par[yr] = xr[xr] +=[yr][yr] =[xr]
        return True

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