290. Word Pattern

Given a pattern and a string str, find if str follows the same pattern.

Here follow means a full match, such that there is a bijection between a letter in pattern and a non-empty word in str.

Examples:

pattern = "abba", str = "dog cat cat dog" should return true.

pattern = "abba", str = "dog cat cat fish" should return false.

pattern = "aaaa", str = "dog cat cat dog" should return false.

pattern = "abba", str = "dog dog dog dog" should return false.

Notes: You may assume pattern contains only lowercase letters, and str contains lowercase letters separated by a single space.

Solution 1.

Keep two hash tables, one map from pattern.charAt[i] to str[i], one map from the other direction. if not exists in either map, update the key-value pair. then compare whether both maps match each other.

public class Solution {
    public boolean wordPattern(String pattern, String str) {
        Map<Character, String> m1 = new HashMap<>();
        Map<String, Character> m2  = new HashMap<>();

        String[] word = str.split(" ");

        if(pattern.length() != word.length) return false;
        for(int i=0;i<pattern.length(); i++){
            char ch = pattern.charAt(i);

            if(!m1.containsKey(ch)) m1.put(ch, word[i]);
            if(!m2.containsKey(word[i])) m2.put(word[i], ch);

            if(!Objects.equals(ch, m2.get(word[i])) || !Objects.equals(word[i], m1.get(ch))){
                return false;
            }
        }

        return true;


    }
}

Solution 2.

There is one more smart solution that take advantage of Map.put()'s return value. and more importantly it utilizes a un-generic map from old java which is not safe.

public boolean wordPattern(String pattern, String str) {
    String[] words = str.split(" ");
    if (words.length != pattern.length())
        return false;
    Map index = new HashMap();
    for (int i=0; i<words.length; ++i)
        if (!Objects.equals(index.put(pattern.charAt(i), i),
                            index.put(words[i], i)))
            return false;
    return true;
}

a bit safer improvement

public class Solution {
    public boolean wordPattern(String pattern, String str) {
        Map<Character, Integer> m1 = new HashMap<>();
        Map<String, Integer> m2 = new HashMap<>();

        String[] words = str.split(" ");
        if(words.length != pattern.length()) return false;

        for(int i=0; i<pattern.length(); i++){
            if(!Objects.equals(m1.put(pattern.charAt(i), i),
                        m2.put(words[i], i))) return false;
        }

        return true;
    }
}