# 839. Similar String Groups

## Description

Two strings, X and Y, are considered similar if either they are identical or we can make them equivalent by swapping at most two letters (in distinct positions) within the string X.

For example, "tars" and "rats" are similar (swapping at positions 0 and 2), and "rats" and "arts" are similar, but "star" is not similar to "tars", "rats", or "arts".

Together, these form two connected groups by similarity: {"tars", "rats", "arts"} and {"star"}.  Notice that "tars" and "arts" are in the same group even though they are not similar.  Formally, each group is such that a word is in the group if and only if it is similar to at least one other word in the group.

We are given a list strs of strings where every string in strs is an anagram of every other string in strs. How many groups are there?

Example 1:

Input: strs = ["tars","rats","arts","star"]
Output: 2


Example 2:

Input: strs = ["omv","ovm"]
Output: 1


Constraints:

• 1 <= strs.length <= 300
• 1 <= strs[i].length <= 300
• strs[i] consists of lowercase letters only.
• All words in strs have the same length and are anagrams of each other.

## Solutions

• class Solution {
private int[] p;

public int numSimilarGroups(String[] strs) {
int n = strs.length;
p = new int[n];
for (int i = 0; i < n; ++i) {
p[i] = i;
}
for (int i = 0; i < n; ++i) {
for (int j = i + 1; j < n; ++j) {
if (check(strs[i], strs[j])) {
p[find(i)] = find(j);
}
}
}
int ans = 0;
for (int i = 0; i < n; ++i) {
if (i == find(i)) {
++ans;
}
}
return ans;
}

private int find(int x) {
if (p[x] != x) {
p[x] = find(p[x]);
}
return p[x];
}

private boolean check(String a, String b) {
int cnt = 0;
for (int i = 0; i < a.length(); ++i) {
if (a.charAt(i) != b.charAt(i)) {
++cnt;
}
}
return cnt <= 2;
}
}

• class Solution {
public:
vector<int> p;

int numSimilarGroups(vector<string>& strs) {
int n = strs.size();
p.resize(n);
for (int i = 0; i < n; ++i) p[i] = i;
for (int i = 0; i < n; ++i)
for (int j = i + 1; j < n; ++j)
if (check(strs[i], strs[j]))
p[find(i)] = find(j);
int ans = 0;
for (int i = 0; i < n; ++i)
if (i == find(i))
++ans;
return ans;
}

bool check(string a, string b) {
int cnt = 0;
for (int i = 0; i < a.size(); ++i)
if (a[i] != b[i])
++cnt;
return cnt <= 2;
}

int find(int x) {
if (p[x] != x) p[x] = find(p[x]);
return p[x];
}
};

• class Solution:
def numSimilarGroups(self, strs: List[str]) -> int:
def find(x):
if p[x] != x:
p[x] = find(p[x])
return p[x]

n, l = len(strs), len(strs[0])
p = list(range(n))
for i in range(n):
for j in range(i + 1, n):
if sum(strs[i][k] != strs[j][k] for k in range(l)) <= 2:
p[find(i)] = find(j)
return sum(i == find(i) for i in range(n))


• func numSimilarGroups(strs []string) int {
n := len(strs)
p := make([]int, n)
for i := range p {
p[i] = i
}
check := func(a, b string) bool {
cnt := 0
for i := range a {
if a[i] != b[i] {
cnt++
}
}
return cnt <= 2
}
var find func(x int) int
find = func(x int) int {
if p[x] != x {
p[x] = find(p[x])
}
return p[x]
}
for i := 0; i < n; i++ {
for j := i + 1; j < n; j++ {
if check(strs[i], strs[j]) {
p[find(i)] = find(j)
}
}
}
ans := 0
for i := 0; i < n; i++ {
if i == find(i) {
ans++
}
}
return ans
}

• class UnionFind {
private p: number[];
private size: number[];

constructor(n: number) {
this.p = Array.from({ length: n }, (_, i) => i);
this.size = Array(n).fill(1);
}

union(a: number, b: number): boolean {
const pa = this.find(a);
const pb = this.find(b);
if (pa === pb) {
return false;
}
if (this.size[pa] > this.size[pb]) {
this.p[pb] = pa;
this.size[pa] += this.size[pb];
} else {
this.p[pa] = pb;
this.size[pb] += this.size[pa];
}
return true;
}

find(x: number): number {
if (this.p[x] !== x) {
this.p[x] = this.find(this.p[x]);
}
return this.p[x];
}
}

function numSimilarGroups(strs: string[]): number {
const n = strs.length;
const m = strs[0].length;
const uf = new UnionFind(n);
let cnt = n;
for (let i = 0; i < n; ++i) {
for (let j = 0; j < i; ++j) {
let diff = 0;
for (let k = 0; k < m; ++k) {
if (strs[i][k] !== strs[j][k]) {
diff++;
}
}
if (diff <= 2 && uf.union(i, j)) {
cnt--;
}
}
}
return cnt;
}