211. Design Add and Search Words Data Structure

Problem

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 <= 25

• word in addWord consists of lowercase English letters.

• word in search consist of '.' or lowercase English letters.

• There will be at most 3 dots in word for search queries.

• At most 10^4 calls will be made to addWord and search.

Solution

class WordDictionary {
    private static class Node {
        char value;
        boolean isEnd;
        Node[] childs;

        Node(char value) {
            this.value = value;
            this.isEnd = false;
            this.childs = new Node[26];
        }

        Node getChild(char ch) {
            return this.childs[ch - 'a'];
        }

        boolean isChild(char ch) {
            return getChild(ch) != null;
        }

        void addChild(char ch) {
            this.childs[ch - 'a'] = new Node(ch);
        }
    }
    // dummy value
    private Node root = new Node('a');

    public WordDictionary() {
        // empty constructor
    }

    public void addWord(String word) {
        Node node = root;
        for (int i = 0; i < word.length(); i++) {
            char ch = word.charAt(i);
            if (!node.isChild(ch)) {
                node.addChild(ch);
            }
            node = node.getChild(ch);
        }
        node.isEnd = true;
    }

    public boolean search(String word) {
        return dfs(this.root, word, 0);
    }

    public boolean dfs(Node root, String word, int index) {
        if (root == null) {
            return false;
        }
        // if reached end of word
        if (index == word.length()) {
            return root.isEnd;
        }
        char ch = word.charAt(index);
        if (ch == '.') {
            for (Node child : root.childs) {
                boolean found = dfs(child, word, index + 1);
                if (found) {
                    return true;
                }
            }
            return false;
        }
        if (!root.isChild(ch)) {
            return false;
        }
        return dfs(root.getChild(ch), word, index + 1);
    }
}

/**
 * Your WordDictionary object will be instantiated and called as such:
 * WordDictionary obj = new WordDictionary();
 * obj.addWord(word);
 * boolean param_2 = obj.search(word);
 */

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).