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))
    
    

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