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 <= 3001 <= strs[i].length <= 300strs[i]consists of lowercase letters only.- All words in
strshave the same length and are anagrams of each other.
Solutions
Solution 1: Union-Find
We can enumerate any two strings $s$ and $t$ in the list of strings. Since $s$ and $t$ are anagrams, if the number of differing characters at corresponding positions between $s$ and $t$ does not exceed $2$, then $s$ and $t$ are similar. We can use the union-find data structure to merge $s$ and $t$. If the merge is successful, the number of similar string groups decreases by $1$.
The final number of similar string groups is the number of connected components in the union-find structure.
Time complexity is $O(n^2 \times (m + \alpha(n)))$, and space complexity is $O(n)$. Here, $n$ and $m$ are the length of the list of strings and the length of the strings, respectively, and $\alpha(n)$ is the inverse Ackermann function, which can be considered a very small constant.
Python3
class UnionFind:
def __init__(self, n):
self.p = list(range(n))
self.size = [1] * n
def find(self, x):
if self.p[x] != x:
self.p[x] = self.find(self.p[x])
return self.p[x]
def union(self, a, b):
pa, pb = self.find(a), self.find(b)
if pa == pb:
return False
if self.size[pa] > self.size[pb]:
self.p[pb] = pa
self.size[pa] += self.size[pb]
else:
self.p[pa] = pb
self.size[pb] += self.size[pa]
return True
class Solution:
def numSimilarGroups(self, strs: List[str]) -> int:
n, m = len(strs), len(strs[0])
uf = UnionFind(n)
for i, s in enumerate(strs):
for j, t in enumerate(strs[:i]):
if sum(s[k] != t[k] for k in range(m)) <= 2 and uf.union(i, j):
n -= 1
return n
Java
class UnionFind {
private final int[] p;
private final int[] size;
public UnionFind(int n) {
p = new int[n];
size = new int[n];
for (int i = 0; i < n; ++i) {
p[i] = i;
size[i] = 1;
}
}
public int find(int x) {
if (p[x] != x) {
p[x] = find(p[x]);
}
return p[x];
}
public boolean union(int a, int b) {
int pa = find(a), pb = find(b);
if (pa == pb) {
return false;
}
if (size[pa] > size[pb]) {
p[pb] = pa;
size[pa] += size[pb];
} else {
p[pa] = pb;
size[pb] += size[pa];
}
return true;
}
}
class Solution {
public int numSimilarGroups(String[] strs) {
int n = strs.length, m = strs[0].length();
UnionFind uf = new UnionFind(n);
int cnt = n;
for (int i = 0; i < n; ++i) {
for (int j = 0; j < i; ++j) {
int diff = 0;
for (int k = 0; k < m; ++k) {
if (strs[i].charAt(k) != strs[j].charAt(k)) {
++diff;
}
}
if (diff <= 2 && uf.union(i, j)) {
--cnt;
}
}
}
return cnt;
}
}
C++
class UnionFind {
public:
UnionFind(int n) {
p = vector<int>(n);
size = vector<int>(n, 1);
iota(p.begin(), p.end(), 0);
}
bool unite(int a, int b) {
int pa = find(a), pb = find(b);
if (pa == pb) {
return false;
}
if (size[pa] > size[pb]) {
p[pb] = pa;
size[pa] += size[pb];
} else {
p[pa] = pb;
size[pb] += size[pa];
}
return true;
}
int find(int x) {
if (p[x] != x) {
p[x] = find(p[x]);
}
return p[x];
}
private:
vector<int> p, size;
};
class Solution {
public:
int numSimilarGroups(vector<string>& strs) {
int n = strs.size(), m = strs[0].size();
int cnt = n;
UnionFind uf(n);
for (int i = 0; i < n; ++i) {
for (int j = 0; j < i; ++j) {
int diff = 0;
for (int k = 0; k < m; ++k) {
diff += strs[i][k] != strs[j][k];
}
if (diff <= 2 && uf.unite(i, j)) {
--cnt;
}
}
}
return cnt;
}
};
Go
type unionFind struct {
p, size []int
}
func newUnionFind(n int) *unionFind {
p := make([]int, n)
size := make([]int, n)
for i := range p {
p[i] = i
size[i] = 1
}
return &unionFind{p, size}
}
func (uf *unionFind) find(x int) int {
if uf.p[x] != x {
uf.p[x] = uf.find(uf.p[x])
}
return uf.p[x]
}
func (uf *unionFind) union(a, b int) bool {
pa, pb := uf.find(a), uf.find(b)
if pa == pb {
return false
}
if uf.size[pa] > uf.size[pb] {
uf.p[pb] = pa
uf.size[pa] += uf.size[pb]
} else {
uf.p[pa] = pb
uf.size[pb] += uf.size[pa]
}
return true
}
func numSimilarGroups(strs []string) int {
n := len(strs)
uf := newUnionFind(n)
for i, s := range strs {
for j, t := range strs[:i] {
diff := 0
for k := range s {
if s[k] != t[k] {
diff++
}
}
if diff <= 2 && uf.union(i, j) {
n--
}
}
}
return n
}
TypeScript
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;
}