LeetCode in Kotlin

890. Find and Replace Pattern

Medium

Given a list of strings words and a string pattern, return a list of words[i] that match pattern. You may return the answer in any order.

A word matches the pattern if there exists a permutation of letters p so that after replacing every letter x in the pattern with p(x), we get the desired word.

Recall that a permutation of letters is a bijection from letters to letters: every letter maps to another letter, and no two letters map to the same letter.

Example 1:

Input: words = [“abc”,”deq”,”mee”,”aqq”,”dkd”,”ccc”], pattern = “abb”

Output: [“mee”,”aqq”]

Explanation: “mee” matches the pattern because there is a permutation {a -> m, b -> e, …}. “ccc” does not match the pattern because {a -> c, b -> c, …} is not a permutation, since a and b map to the same letter.

Example 2:

Input: words = [“a”,”b”,”c”], pattern = “a”

Output: [“a”,”b”,”c”]

Constraints:

• 1 <= pattern.length <= 20
• 1 <= words.length <= 50
• words[i].length == pattern.length
• pattern and words[i] are lowercase English letters.

Solution

class Solution {
    fun findAndReplacePattern(words: Array<String>, pattern: String): List<String> {
        val finalans: MutableList<String> = ArrayList()
        if (pattern.length == 1) {
            finalans.addAll(words)
            return finalans
        }
        for (word in words) {
            val check = CharArray(26)
            check.fill('1')
            val ans: HashMap<Char, Char> = HashMap()
            for (j in word.indices) {
                val pat = pattern[j]
                val wor = word[j]
                if (ans.containsKey(pat)) {
                    if (ans[pat] == wor) {
                        if (j == word.length - 1) {
                            finalans.add(word)
                        }
                    } else {
                        break
                    }
                } else {
                    if (j == word.length - 1 && check[wor.code - 'a'.code] == '1') {
                        finalans.add(word)
                    }
                    if (check[wor.code - 'a'.code] != '1' && check[wor.code - 'a'.code] != pat) {
                        break
                    }
                    if (check[wor.code - 'a'.code] == '1') {
                        ans[pat] = wor
                        check[wor.code - 'a'.code] = pat
                    }
                }
            }
        }
        return finalans
    }
}

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