Leetcode211-Design Add and Search Words Data Structure

Description

Design a data structure that supports adding new words and finding if a string matches any previously added string.

Implement the WordDictionary class:

• WordDictionary() Initializes the object.
• void addWord(word) Adds word to the data structure, it can be matched later.
• bool search(word) Returns true if there is any string in the data structure that matches word or false otherwise. word may contain dots ‘.’ where dots can be matched with any letter.

Example

Input
["WordDictionary","addWord","addWord","addWord","search","search","search","search"]
[[],["bad"],["dad"],["mad"],["pad"],["bad"],[".ad"],["b.."]]
Output
[null,null,null,null,false,true,true,true]

Explanation
WordDictionary wordDictionary = new WordDictionary();
wordDictionary.addWord("bad");
wordDictionary.addWord("dad");
wordDictionary.addWord("mad");
wordDictionary.search("pad"); // return False
wordDictionary.search("bad"); // return True
wordDictionary.search(".ad"); // return True
wordDictionary.search("b.."); // return True

Constraints:

• 1 <= word.length <= 500
• word in addWord consists lower-case English letters.
• word in search consist of ‘.’ or lower-case English letters.
• At most 50000 calls will be made to addWord and search.

Solution

Solution 1: HashMap, O(MN^2)

// O(N^2 * M)
class WordDictionary {
    Map<Integer, Set<String>> d;

    /** Initialize your data structure here. */
    public WordDictionary() {
        d = new HashMap();
    }

    /** Adds a word into the data structure. */
    public void addWord(String word) {
        int m = word.length();
        if (!d.containsKey(m)) {
            d.put(m, new HashSet());
        }
        d.get(m).add(word);
    }

    /** Returns if the word is in the data structure. A word could contain the dot character '.' to represent any one letter. */
    public boolean search(String word) {
        int m = word.length();
        if (d.containsKey(m)) {
            for (String w : d.get(m)) {
                int i = 0;
                while ((i < m) && (w.charAt(i) == word.charAt(i) || word.charAt(i) == '.')) {
                    i++;
                }
                if (i == m) return true;
            }
        }
        return false;
    }
}

Solution 2: Trie Tree, O(MN) for defined words without dots, O(NM^26) for undefined words

class TrieNode {
    Map<Character, TrieNode> children = new HashMap<>();
    boolean isWord = false;
    
    public TrieNode(){
    }
}

class WordDictionary {
    private TrieNode node;
    
    public WordDictionary() {
        node = new TrieNode();
    }
    
    public void addWord(String word) {
        TrieNode cur = node;
        for (char ch: word.toCharArray()){
            if (!cur.children.containsKey(ch)){
                cur.children.put(ch, new TrieNode());
            }
            cur = cur.children.get(ch);
        }
        cur.isWord = true;
    }
    
    private boolean searchTrieTree(TrieNode node, String word){
        for (int i = 0; i < word.length(); i++){
            char ch = word.charAt(i);
            if (!node.children.containsKey(ch)){
                if (ch == '.'){
                    for (char c: node.children.keySet()) {
                        TrieNode child = node.children.get(c);
                        if (searchTrieTree(child, word.substring(i + 1))){
                            return true;
                        }
                    }
                }
                return false;
            }
            else{
                node = node.children.get(ch);
            }
        }
        return node.isWord;
    } 
    
    public boolean search(String word) {
        return searchTrieTree(node,word);
    }
}

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