839. Similar String Groups

package LeetCode_839

import java.util.*
import kotlin.collections.HashSet

/**
 * 839. Similar String Groups
 * https://leetcode.com/problems/similar-string-groups/
 * Two strings X and Y are similar if we can swap two letters (in different positions) of X, so that it equals Y.
 * Also two strings X and Y are similar if they are equal.
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. 1 <= strs.length <= 100
2. 1 <= strs[i].length <= 1000
3. sum(strs[i].length) <= 2 * 10^4
4. strs[i] consists of lowercase letters only.
5. All words in strs have the same length and are anagrams of each other.
 * */
class Solution {
    /*
    * solution: BFS, do bfs for each string to check current and next if similar,
    * Time:O(n^2*l), l is length of str, Space:O(n),
    * */
    fun numSimilarGroups(strs: Array<String>): Int {
        if (strs.size < 2) {
            return strs.size
        }
        var group = 0
        val visited = HashSet<String>()
        val queue = LinkedList<String>()
        for (str in strs) {
            if (visited.contains(str)) {
                continue
            }
            visited.add(str)
            queue.offer(str)
            group++
            while (queue.isNotEmpty()) {
                val cur = queue.poll()
                //cur string compare with next string
                for (item in strs) {
                    if (visited.contains(item)) {
                        continue
                    }
                    var diff = 0
                    //if the number of different char equals to 2, is similar;
                    //all word have the same length
                    for (i in cur.indices) {
                        if (cur[i] != item[i]) {
                            diff++
                        }
                    }
                    //put similar string into queue to check next level
                    if (diff == 2) {
                        visited.add(item)
                        queue.offer(item)
                    }
                }
            }
        }

        return group
    }
}
原文地址:https://www.cnblogs.com/johnnyzhao/p/14280042.html