# 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.

Constraints

n ≤ 10,000

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

Example 1

Input

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

Output

2

Explanation

There are two mutation groups:

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

["TTTT", "TTTG"]```

### Solution :

```                        ```Solution in C++ :

class UnionFind {
private:
vector<int> parents, rank;

public:
UnionFind(int n) {
parents.resize(n);
rank.resize(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;
rank[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)
return;
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.mp = {}
self.par = []
self.sz = []

def find(self, x):
try:
i = self.mp[x]
except:
self.mp[x] = i = len(self.mp)
self.par.append(i)
self.sz.append(1)
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 self.sz[xr] < self.sz[yr]:
xr, yr = yr, xr
self.par[yr] = xr
self.sz[xr] += self.sz[yr]
self.sz[yr] = self.sz[xr]
return True```
```

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