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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

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.

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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

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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 &&