211. Design Add and Search Words Data Structure

Problem Statement

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.

Click to open Hints

You should be familiar with how a Trie works. If not, please work on this problem: Implement Trie (Prefix Tree) first.

Solution:

package main

type WordDictionary struct {
    children [26]* WordDictionary
    isWord bool

}


func Constructor() WordDictionary {
    return WordDictionary{}
}


func (this *WordDictionary) AddWord(word string)  {
    root := this

    for _,item := range word{
        char := item - 'a'

        if root.children[char] == nil {
            root.children[char] = &WordDictionary{}
        }
        root = root.children[char]
    }
    root.isWord = true

}


func (this *WordDictionary) Search(word string) bool {
    root := this
    for i,item := range word{
        if word[i] == '.'{
            for _,item := range root.children{
                if item != nil && item.Search(word[i+1:]) {
                    return true
                }
            }

            return false
        }

        char := item - 'a'
        if root.children[char] == nil{
            return false
        }
        root = root.children[char]
    }

    return root.isWord
}

211. Design Add and Search Words Data Structure #

Problem Statement #

Solution: #

... #

211. Design Add and Search Words Data Structure

Problem Statement

Solution:

...