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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0801_0900.s0890_find_and_replace_pattern;

// #Medium #Array #String #Hash_Table #2022_03_28_Time_1_ms_(97.13%)_Space_42.3_MB_(86.23%)

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.List;

/**
 * 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.
**/
@SuppressWarnings("java:S135")
public class Solution {
    public List findAndReplacePattern(String[] words, String pattern) {
        List finalans = new ArrayList<>();
        if (pattern.length() == 1) {
            Collections.addAll(finalans, words);
            return finalans;
        }
        for (String word : words) {
            char[] check = new char[26];
            Arrays.fill(check, '1');
            HashMap ans = new HashMap<>();
            for (int j = 0; j < word.length(); j++) {
                char pat = pattern.charAt(j);
                char wor = word.charAt(j);
                if (ans.containsKey(pat)) {
                    if (ans.get(pat) == wor) {
                        if (j == word.length() - 1) {
                            finalans.add(word);
                        }
                    } else {
                        break;
                    }
                } else {
                    if (j == word.length() - 1 && check[wor - 'a'] == '1') {
                        finalans.add(word);
                    }
                    if (check[wor - 'a'] != '1' && check[wor - 'a'] != pat) {
                        break;
                    }
                    if (check[wor - 'a'] == '1') {
                        ans.put(pat, wor);
                        check[wor - 'a'] = pat;
                    }
                }
            }
        }
        return finalans;
    }
}