[📝LeetCode #290] Word Pattern

#leetcode #java #100daysofcode #hashtable

🎀 The Problem

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.

Example:

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

👩‍💻 My Answer

class Solution { public boolean wordPattern(String pattern, String s) { HashMap map = new HashMap(); int count = 0, wordcount = 0; String word = ""; for (int i = 0; i < s.length(); i++) { char letter = s.charAt(i); if (count == pattern.length()) return false; if (letter != ' ') word += letter; if (i == s.length() - 1 || word != "" && letter == ' ' ){ char key = pattern.charAt(count); if (!map.containsKey(key)) if (!map.containsValue(word)) { map.put(key, word); wordcount++; } else return false; else if (map.containsKey(key)) if (!map.get(key).equals(word)) return false; else wordcount++; count++; word = ""; } } if (wordcount != pattern.length()) return false; return true; } }

Pro & Con

✖️ Runtime & Memory

💋 Ideal Answer

Approach

Ok, I did not know there is a library that can split words by " " and store them into the String[]. This lack of knowledge costs me so many lines and runtime.

Here is my new code that BEATS 100%.

New Code

class Solution { public boolean wordPattern(String pattern, String s) { String[] words = s.split(" "); if (words.length != pattern.length()) return false; HashMap map = new HashMap(); for (int i = 0; i < words.length; i++) { char key = pattern.charAt(i); if (!map.containsKey(key) && !map.containsValue(words[i])) map.put(key, words[i]); else if (map.containsKey(key) && map.get(key).equals(words[i])) continue; else return false; } return true; } }