Question

Formatted question description: https://leetcode.ca/all/290.html

 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.

 Example 1:

 Input: pattern = "abba", str = "dog cat cat dog"
 Output: true

 Example 2:

 Input:pattern = "abba", str = "dog cat cat fish"
 Output: false

 Example 3:

 Input: pattern = "aaaa", str = "dog cat cat dog"
 Output: false

 Example 4:

 Input: pattern = "abba", str = "dog dog dog dog"
 Output: false

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

Algorithm

If the one-to-one mapping cannot be established, return false directly.

It also returns false when the mapping values of the two HashMaps are not the same.

Code

Java

import java.util.HashMap;
import java.util.Map;

public class Word_Pattern {


    public class Solution {
        public boolean wordPattern(String pattern, String str) {
            if (pattern == null || pattern.length() == 0 || str == null || str.length() == 0) {
                return false;
            }

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

            if (pattern.length() != tokens.length) {
                return false;
            }

            Map<String, Character> inverseMap = new HashMap<>();
            Map<Character, String> map = new HashMap<>();

            for (int i = 0; i < pattern.length(); i++) {
                char c = pattern.charAt(i);
                String token = tokens[i];

                // Check the one-one mapping
                if (!map.containsKey(c) && !inverseMap.containsKey(token)) {
                    map.put(c, token);
                    inverseMap.put(token, c);
                } else if (map.containsKey(c) && inverseMap.containsKey(token)) {
                    String tokenToCheck = map.get(c);
                    char charToCheck = inverseMap.get(token);

                    if (!tokenToCheck.equals(token) || c != charToCheck) {
                        return false;
                    }
                } else { // @note: this is for different length of pattern and str
                    return false;
                }
            }

            return true;
        }
    }
}

// OJ: https://leetcode.com/problems/word-pattern/
// Time: O(M + N)
// Space: O(M + N)
class Solution {
public:
    bool wordPattern(string pattern, string str) {
        unordered_map<char, string> m;
        unordered_map<string, char> r;
        istringstream ss(str);
        string word;
        for (char c : pattern) {
            if (!(ss >> word)) return false;
            if ((m.count(c) && m[c] != word)
                || r.count(word) && r[word] != c) return false;
            m[c] = word;
            r[word] = c;
        }
        return ss.eof();
    }
};

class Solution(object):
  def wordPattern(self, pattern, str):
    """
    :type pattern: str
    :type str: str
    :rtype: bool
    """
    str = str.split()
    a = zip(pattern, str)
    print
    a
    return len(pattern) == len(str) and len(set(a)) == len(set(pattern)) == len(set(str))

290 - Word Pattern

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