Question

Formatted question description: https://leetcode.ca/all/211.html

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 2 dots in word for search queries.
At most 10⁴ calls will be made to addWord and search.

Algorithm

With the structure of the Trie, the only difference is that the search function needs to be rewritten, compared with Trie problem.

Because the . in this question can replace any character, once there is a ., all subtrees need to be searched. As long as one returns true, the entire search function returns true.

Typical DFS problem.

Code

public class Add_and_Search_Word_Data_structure_design {

    class WordDictionary {

        Trie trie;

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

        /** Adds a word into the data structure. */
        public void addWord(String word) {
            trie.insert(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) {
            return trie.search(word);
        }
    }

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

    class TrieNode {

        // R children to node children
        public TrieNode[] children;

        private final int R = 26;

        private boolean isEnd;

        public TrieNode() {
            children = new TrieNode[R];
        }

        public boolean containsKey(char ch) {
            return children[ch -'a'] != null;
        }
        public TrieNode get(char ch) {
            return children[ch -'a'];
        }
        public void put(char ch, TrieNode node) {
            children[ch -'a'] = node;
        }
        public void setEnd() {
            isEnd = true;
        }
        public boolean isEnd() {
            return isEnd;
        }
    }

    class Trie {
        private TrieNode root;

        public Trie() {
            root = new TrieNode();
        }

        // Inserts a word into the trie.
        public void insert(String word) {
            TrieNode node = root;

            // @note:@memorize: iteration is better than recursion during trie building
            for (int i = 0; i < word.length(); i++) {
                char currentChar = word.charAt(i);
                if (!node.containsKey(currentChar)) {
                    node.put(currentChar, new TrieNode());
                }
                node = node.get(currentChar);
            }
            node.setEnd();
        }

        // search a prefix or whole key in trie and
        // returns the node where search ends
        public boolean search(String word) {

            return dfs(word, root);

        }

        private boolean dfs(String word, TrieNode node) {

            if (node == null) {
                return false;
            }

            if (word.length() == 0) {
                return node.isEnd();
            }

            char curLetter = word.charAt(0);
            if (curLetter == '.') {
                for (int i = 0; i < 26; i++) {
                    char c = (char)('a' + i);
                    if (dfs(word.substring(1), node.get(c))) {
                        return true;
                    }
                }
            } else if (node.containsKey(curLetter)){
                return dfs(word.substring(1), node.get(curLetter));
            }

            return false;
        }

    }


    public class WordDictionary_bucket {

        // bucket: word length -> list of words
        Map<Integer, List<String>> map = new HashMap<Integer, List<String>>();

        // Adds a word into the data structure.
        public void addWord(String word) {
            int index = word.length();
            if(!map.containsKey(index)){
                List<String> list = new ArrayList<String>();
                list.add(word);
                map.put(index, list);
            }else{
                map.get(index).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 index = word.length();
            if(!map.containsKey(index)){
                return false;
            }
            List<String> list = map.get(index);
            if(isWords(word)){
                return list.contains(word);
            }
            for(String s : list){
                if(isSame(s, word)){
                    return true;
                }
            }
            return false;
        }

        boolean isWords(String s){
            for(int i = 0; i < s.length(); i++){

                // @note: api isLetter()
                if(!Character.isLetter(s.charAt(i))){
                    return false;
                }
            }
            return true;
        }

        boolean isSame(String a, String search){
            if(a.length() != search.length()){
                return false;
            }
            for(int i = 0; i < a.length(); i++){
                if(search.charAt(i) != '.' && search.charAt(i) != a.charAt(i)){
                    return false;
                }
            }
            return true;
        }
    }

}


############

class Trie {
    Trie[] children = new Trie[26];
    boolean isEnd;
}

class WordDictionary {
    private Trie trie;

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

    public void addWord(String word) {
        Trie node = trie;
        for (char c : word.toCharArray()) {
            int idx = c - 'a';
            if (node.children[idx] == null) {
                node.children[idx] = new Trie();
            }
            node = node.children[idx];
        }
        node.isEnd = true;
    }

    public boolean search(String word) {
        return search(word, trie);
    }

    private boolean search(String word, Trie node) {
        for (int i = 0; i < word.length(); ++i) {
            char c = word.charAt(i);
            int idx = c - 'a';
            if (c != '.' && node.children[idx] == null) {
                return false;
            }
            if (c == '.') {
                for (Trie child : node.children) {
                    if (child != null && search(word.substring(i + 1), child)) {
                        return true;
                    }
                }
                return false;
            }
            node = node.children[idx];
        }
        return node.isEnd;
    }
}

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

// OJ: https://leetcode.com/problems/add-and-search-word-data-structure-design/
// Time:
//      WordDictionary: O(1)
//      addWord: O(W)
//      search: O(26^W)
// Space: O(NW)
struct TrieNode {
    TrieNode *next[26] = {};
    bool end = false;
};
class WordDictionary {
    TrieNode root;
    bool dfs(TrieNode *node, string &word, int i) {
        if (!node) return false;
        if (i == word.size()) return node->end;
        if (word[i] != '.') return dfs(node->next[word[i] - 'a'], word, i + 1);
        for (int j = 0; j < 26; ++j) {
            if (dfs(node->next[j], word, i + 1)) return true;
        }
        return false;
    }
public:
    void addWord(string word) {
        auto node = &root;
        for (char c : word) {
            if (!node->next[c - 'a']) node->next[c - 'a'] = new TrieNode();
            node = node->next[c - 'a'];
        }
        node->end = true;
    }
    bool search(string word) {
        return dfs(&root, word, 0);
    }
};

class Trie:
    def __init__(self):
        self.children = [None] * 26
        self.is_end = False


class WordDictionary:
    def __init__(self):
        self.trie = Trie()

    def addWord(self, word: str) -> None:
        node = self.trie
        for c in word:
            idx = ord(c) - ord('a')
            if node.children[idx] is None:
                node.children[idx] = Trie()
            node = node.children[idx]
        node.is_end = True

    def search(self, word: str) -> bool:
        def search(word, node):
            for i in range(len(word)):
                c = word[i]
                idx = ord(c) - ord('a')
                if c != '.' and node.children[idx] is None:
                    return False
                if c == '.':
                    for child in node.children:
                        if child is not None and search(word[i + 1 :], child):
                            return True
                    return False
                node = node.children[idx]
            return node.is_end

        return search(word, self.trie)


# Your WordDictionary object will be instantiated and called as such:
# obj = WordDictionary()
# obj.addWord(word)
# param_2 = obj.search(word)

############

class TrieNode:
  def __init__(self):
    self.neighbours = {}
    self.isWord = False


class Trie:
  def __init__(self):
    self.root = TrieNode()

  def addWord(self, word):
    root = self.root
    for i in range(0, len(word)):
      c = word[i]
      if c in root.neighbours:
        root = root.neighbours[c]
      else:
        newnode = TrieNode()
        root.neighbours[c] = newnode
        root = root.neighbours[c]
    root.isWord = True


class WordDictionary:
  def __init__(self):
    self.trie = Trie()
    self.cache = set([])

  def addWord(self, word):
    self.trie.addWord(word)
    self.cache.add(word)

  def search(self, word):
    if word in self.cache:
      return True

    def dfsHelper(root, word, index):
      if not root:
        return False

      if len(word) == index:
        if root.isWord:
          return True
        return False

      if word[index] != ".":
        if dfsHelper(root.neighbours.get(word[index], None), word, index + 1):
          return True
      else:
        for nbr in root.neighbours:
          if dfsHelper(root.neighbours[nbr], word, index + 1):
            return True
      return False

    return dfsHelper(self.trie.root, word, 0)

type WordDictionary struct {
	root *trie
}

func Constructor() WordDictionary {
	return WordDictionary{new(trie)}
}

func (this *WordDictionary) AddWord(word string) {
	this.root.insert(word)
}

func (this *WordDictionary) Search(word string) bool {
	n := len(word)

	var dfs func(int, *trie) bool
	dfs = func(i int, cur *trie) bool {
		if i == n {
			return cur.isEnd
		}
		c := word[i]
		if c != '.' {
			child := cur.children[c-'a']
			if child != nil && dfs(i+1, child) {
				return true
			}
		} else {
			for _, child := range cur.children {
				if child != nil && dfs(i+1, child) {
					return true
				}
			}
		}
		return false
	}

	return dfs(0, this.root)
}

type trie struct {
	children [26]*trie
	isEnd    bool
}

func (t *trie) insert(word string) {
	cur := t
	for _, c := range word {
		c -= 'a'
		if cur.children[c] == nil {
			cur.children[c] = new(trie)
		}
		cur = cur.children[c]
	}
	cur.isEnd = true
}

/**
 * Your WordDictionary object will be instantiated and called as such:
 * obj := Constructor();
 * obj.AddWord(word);
 * param_2 := obj.Search(word);
 */

using System.Collections.Generic;
using System.Linq;

class TrieNode {
    public bool IsEnd { get; set; }
    public TrieNode[] Children { get; set; }
    public TrieNode() {
        Children = new TrieNode[26];
    }
}

public class WordDictionary {
    private TrieNode root;

    public WordDictionary() {
        root = new TrieNode();
    }

    public void AddWord(string word) {
        var node = root;
        for (var i = 0; i < word.Length; ++i)
        {
            TrieNode nextNode;
            var index = word[i] - 'a';
            nextNode = node.Children[index];
            if (nextNode == null)
            {
                nextNode = new TrieNode();
                node.Children[index] = nextNode;
            }
            node = nextNode;
        }
        node.IsEnd = true;
    }

    public bool Search(string word) {
        var queue = new Queue<TrieNode>();
        queue.Enqueue(root);
        for (var i = 0; i < word.Length; ++i)
        {
            var count = queue.Count;
            while (count-- > 0)
            {
                var node = queue.Dequeue();
                if (word[i] == '.')
                {
                    foreach (var nextNode in node.Children)
                    {
                        if (nextNode != null)
                        {
                            queue.Enqueue(nextNode);
                        }
                    }
                }
                else
                {
                    var nextNode = node.Children[word[i] - 'a'];
                    if (nextNode != null)
                    {
                        queue.Enqueue(nextNode);
                    }
                }
            }
        }
        return queue.Any(n => n.IsEnd);
    }
}

211 - Design Add and Search Words Data Structure

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