211. 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 <= 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.

Solutions

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.

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);
 */

class trie {
public:
    vector<trie*> children;
    bool is_end;

    trie() {
        children = vector<trie*>(26, nullptr);
        is_end = false;
    }

    void insert(const string& word) {
        trie* cur = this;
        for (char c : word) {
            c -= 'a';
            if (cur->children[c] == nullptr) {
                cur->children[c] = new trie;
            }
            cur = cur->children[c];
        }
        cur->is_end = true;
    }
};

class WordDictionary {
private:
    trie* root;

public:
    WordDictionary()
        : root(new trie) {}

    void addWord(string word) {
        root->insert(word);
    }

    bool search(string word) {
        return dfs(word, 0, root);
    }

private:
    bool dfs(const string& word, int i, trie* cur) {
        if (i == word.size()) {
            return cur->is_end;
        }
        char c = word[i];
        if (c != '.') {
            trie* child = cur->children[c - 'a'];
            if (child != nullptr && dfs(word, i + 1, child)) {
                return true;
            }
        } else {
            for (trie* child : cur->children) {
                if (child != nullptr && dfs(word, i + 1, child)) {
                    return true;
                }
            }
        }
        return false;
    }
};

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

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)

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

211. Design Add and Search Words Data Structure

Description

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211. Design Add and Search Words Data Structure

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