LeetCode in Kotlin

839. Similar String Groups

Hard

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:

Solution

import java.util.LinkedList
import java.util.Queue

class Solution {
    fun numSimilarGroups(strs: Array<String>): Int {
        val visited = BooleanArray(strs.size)
        var res = 0
        for (i in strs.indices) {
            if (!visited[i]) {
                res++
                bfs(i, visited, strs)
            }
        }
        return res
    }

    private fun bfs(i: Int, visited: BooleanArray, strs: Array<String>) {
        val qu: Queue<String> = LinkedList()
        qu.add(strs[i])
        visited[i] = true
        while (qu.isNotEmpty()) {
            val s = qu.poll()
            for (j in strs.indices) {
                if (visited[j]) {
                    continue
                }
                if (isSimilar(s, strs[j])) {
                    visited[j] = true
                    qu.add(strs[j])
                }
            }
        }
    }

    private fun isSimilar(s1: String, s2: String): Boolean {
        var c1: Char? = null
        var c2: Char? = null
        var mismatchCount = 0
        for (i in s1.indices) {
            if (s1[i] == s2[i]) {
                continue
            }
            mismatchCount++
            if (c1 == null) {
                c1 = s1[i]
                c2 = s2[i]
            } else if (s2[i] != c1 || s1[i] != c2) {
                return false
            } else if (mismatchCount > 2) {
                return false
            }
        }
        return true
    }
}
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