Tags

  • AWS (7)
  • Apigee (3)
  • ArchLinux (5)
  • Array (6)
  • Backtracking (6)
  • BinarySearch (6)
  • C++ (19)
  • CI&CD (3)
  • Calculus (2)
  • DesignPattern (43)
  • DisasterRecovery (1)
  • Docker (8)
  • DynamicProgramming (20)
  • FileSystem (11)
  • Frontend (2)
  • FunctionalProgramming (1)
  • GCP (1)
  • Gentoo (6)
  • Git (15)
  • Golang (1)
  • Graph (10)
  • GraphQL (1)
  • Hardware (1)
  • Hash (1)
  • Kafka (1)
  • LinkedList (13)
  • Linux (27)
  • Lodash (2)
  • MacOS (3)
  • Makefile (1)
  • Map (5)
  • MathHistory (1)
  • MySQL (21)
  • Neovim (10)
  • Network (66)
  • Nginx (6)
  • Node.js (33)
  • OpenGL (6)
  • PriorityQueue (1)
  • ProgrammingLanguage (9)
  • Python (10)
  • RealAnalysis (20)
  • Recursion (3)
  • Redis (1)
  • RegularExpression (1)
  • Ruby (19)
  • SQLite (1)
  • Sentry (3)
  • Set (4)
  • Shell (3)
  • SoftwareEngineering (12)
  • Sorting (2)
  • Stack (4)
  • String (2)
  • SystemDesign (13)
  • Terraform (2)
  • Tree (24)
  • Trie (2)
  • TwoPointers (16)
  • TypeScript (3)
  • Ubuntu (4)

    • Home

    [LeetCode 211] Design Add and Search Words Data Structure

    Published Jun 23, 2022 [  Trie  ]

    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.

    Thoughts

    • We can consider using Trie for word search problem

    TypeScript

    class TrieNode {
    children: Map<string, TrieNode>
    isEnd: boolean
    
    constructor(){
        this.children = new Map<string, TrieNode>()
        this.isEnd = false;
    }
    
}

class WordDictionary {
    root: TrieNode;
    
    constructor() {
        this.root = new TrieNode()
    }

    addWord(word: string): void {
        let cur: TrieNode = this.root;
        for(const c of word){
            if(!cur.children.has(c)) cur.children.set(c, new TrieNode())
            cur = cur.children.get(c)
        }
        cur.isEnd = true;
    }

    search(word: string): boolean {
        let res: boolean = false;
        const LEN: number = word.length
        let cur: TrieNode = this.root
        
        const exec = (i: number, node: TrieNode) => {
            if(res === true) return;
            if(i === LEN){
                if(node.isEnd === true) res = true
                return;
            }
            
            if(word[i] === '.') {
                for(const c of node.children.keys()){
                    exec(i + 1, node.children.get(c))
                }
            } else if(node.children.has(word[i])){
                exec(i + 1, node.children.get(word[i]))
            }
        }
        
        exec(0, cur)
        return res;
    }
}

    Reference