839 Similar String Groups

839. Similar String Groups

1. Question

Two stringsX andY are similar if we can swap two letters (in different positions) ofX, so that it equalsY.

For example,"tars" and"rats" are similar (swapping at positions0and2), 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 listAof strings. Every string inAis an anagram of every other string inA. How many groups are there?

Example 1:

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

Note:

1. A.length <= 2000

2. A[i].length <= 1000

3. A.length * A[i].length <= 20000

4. All words inAconsist of lowercase letters only.

5. All words inAhave the same length and are anagrams of each other.

6. The judging time limit has been increased for this question.

2. Implementation

(1) Union Find

思路: Union Find的思想比较直接，两个string只要是similar的就直接union.需要注意的是isSimilar()里，按题目要求应该是diff等于2属于similar，但为了避免输入数组中的相同string，我们把相同string也union在一起。

class Solution {
    public int numSimilarGroups(String[] A) {
        if (A.length <= 1) {
            return A.length;
        }

        UnionFind uf = new UnionFind(A.length);

        for (int i = 0; i < A.length - 1; i++) {
            for (int j = i + 1; j < A.length; j++) {
                if (isSimilar(A[i], A[j])) {
                    uf.union(i, j);
                }
            }
        }
        return uf.count;
    }

    public boolean isSimilar(String s1, String s2) {
        if (s1.length() != s2.length()) {
            return false;
        }

        int diff = 0;
        for (int i = 0; i < s1.length(); i++) {
            if (s1.charAt(i) != s2.charAt(i)) {
                ++diff;
            }
        }
        return diff == 0 || diff == 2;
    }

    class UnionFind {
        int[] parents;
        int[] size;
        int count;

        public UnionFind(int n) {
            parents = new int[n];
            size = new int[n];
            count = n;

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

        public int find(int id) {
            while (id != parents[id]) {
                id = parents[id];
            }
            return id;
        }

        public void union(int i, int j) {
            int rootI = find(i);
            int rootJ = find(j);

            if (rootI == rootJ) return;

            if (size[rootI] < size[rootJ]) {
                parents[rootI] = rootJ;
                size[rootJ] += size[rootI];
            }
            else {
                parents[rootJ] = rootI;
                size[rootI] += size[rootJ];
            }
            --count;
        }
    }
}

3. Time & Space Complexity

Union Find: 时间复杂度O(n^2 * L), 空间复杂度O(n^2)