Design Add and Search Words Data Structure LeetCode Solution

Design Add and Search Words Data Structure 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 104 calls will be made to addWord and search.

Design Add and Search Words Data Structure Solutions

✅Time: 
✅Space:

C++

struct TrieNode {
  vector<TrieNode*> children;
  bool isWord = false;
  TrieNode() : children(26) {}
  ~TrieNode() {
    for (TrieNode* child : children)
      delete child;
  }
};

class WordDictionary {
 public:
  void addWord(const string& word) {
    TrieNode* node = &root;
    for (const char c : word) {
      const int i = c - 'a';
      if (!node->children[i])
        node->children[i] = new TrieNode;
      node = node->children[i];
    }
    node->isWord = true;
  }

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

 private:
  TrieNode root;

  bool dfs(const string& word, int s, TrieNode* node) {
    if (s == word.length())
      return node->isWord;
    if (word[s] != '.') {
      TrieNode* next = node->children[word[s] - 'a'];
      return next ? dfs(word, s + 1, next) : false;
    }

    // word[s] == '.' -> search all 26 children
    for (int i = 0; i < 26; ++i)
      if (node->children[i] && dfs(word, s + 1, node->children[i]))
        return true;

    return false;
  }
};

Java

 class TrieNode {
  public TrieNode[] children = new TrieNode[26];
  public boolean isWord = false;
}

class WordDictionary {
  public void addWord(String word) {
    TrieNode node = root;
    for (final char c : word.toCharArray()) {
      final int i = c - 'a';
      if (node.children[i] == null)
        node.children[i] = new TrieNode();
      node = node.children[i];
    }
    node.isWord = true;
  }

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

  private TrieNode root = new TrieNode();

  private boolean dfs(String word, int s, TrieNode node) {
    if (s == word.length())
      return node.isWord;
    if (word.charAt(s) != '.') {
      TrieNode next = node.children[word.charAt(s) - 'a'];
      return next == null ? false : dfs(word, s + 1, next);
    }

    // word.charAt(s) == '.' -> search all 26 children
    for (int i = 0; i < 26; ++i)
      if (node.children[i] != null && dfs(word, s + 1, node.children[i]))
        return true;

    return false;
  }
}

Python

 class WordDictionary:
  def __init__(self):
    self.root = {}

  def addWord(self, word: str) -> None:
    node = self.root
    for c in word:
      if c not in node:
        node[c] = {}
      node = node[c]
    node['isWord'] = True

  def search(self, word: str) -> bool:
    return self.dfs(word, 0, self.root)

  def dfs(self, word: str, s: int, node: dict) -> bool:
    if s == len(word):
      return 'isWord' in node
    if word[s] != '.':
      if word[s] in node:
        return self.dfs(word, s + 1, node[word[s]])
      return False

    for c in string.ascii_lowercase:
      if c in node and self.dfs(word, s + 1, node[c]):
        return True

    return False

Watch Tutorial

Checkout more Solutions here