290. Word Pattern

Given a pattern and a string s, find if s 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 s. Specifically:

• Each letter in pattern maps to exactly one unique word in s.
• Each unique word in s maps to exactly one letter in pattern.
• No two letters map to the same word, and no two words map to the same letter.

Example 1:

Input: pattern = "abba", s = "dog cat cat dog"

Output: true

Explanation:

The bijection can be established as:

• 'a' maps to "dog".
• 'b' maps to "cat".

Example 2:

Input: pattern = "abba", s = "dog cat cat fish"

Output: false

Example 3:

Input: pattern = "aaaa", s = "dog cat cat dog"

Output: false

Constraints:

• 1 <= pattern.length <= 300
• pattern contains only lower-case English letters.
• 1 <= s.length <= 3000
• s contains only lowercase English letters and spaces ' '.
• s does not contain any leading or trailing spaces.
• All the words in s are separated by a single space.

Solution:

class Solution {
    public boolean wordPattern(String pattern, String s) {
        Map<Character, String> mapPattern = new HashMap<>();
        Map<String, Character> mapS = new HashMap<>();
        int n = pattern.length();
        String[] sArray = s.split(" ");
        if (sArray.length != n){
            return false;
        }

        for (int i = 0; i < n; i++){
            char curPattern = pattern.charAt(i);
            String curS = sArray[i];

            if (!mapPattern.containsKey(curPattern) && !mapS.containsKey(curS)){
                mapPattern.put(curPattern, curS);
                mapS.put(curS, curPattern);
            }else if (mapPattern.containsKey(curPattern) && mapS.containsKey(curS)){
                String curMapPattern = mapPattern.get(curPattern);
                char curMapS = mapS.get(curS);
                if (!curMapPattern.equals(curS) || curMapS != curPattern){
                    return false;
                }
            }else{
                return false;
            }
        }
        return true;
    }
}

// TC: O(n)
// SC: O(n)