211. 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.
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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 <= 500
  • word in addWord consists lower-case English letters.
  • word in search consist of ‘.’ or lower-case English letters.
  • At most 50000 calls will be made to addWord and search.

Solution

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class WordDictionary {
    static int R = 256;
    Trie trie = new Trie(); 

    /** Initialize your data structure here. */
    public WordDictionary() {
        
    }
    
    public void addWord(String word) {
        Trie node = trie;
        for (char c: word.toCharArray()) {
            Trie curr = node.next[c];
            if (curr == null) {
                curr = new Trie();
            }
            node.next[c] = curr;
            node = curr;
        }
        node.isWord = true;
    }
    
    public boolean search(String word) {
        return patternSearch(word, trie, 0);
    }
    
    boolean patternSearch(String word, Trie node, int index) {
         if (node == null) return false;
         if (index == word.length()) return node.isWord;
         char c = word.charAt(index);             
         if (c == '.') {
             for (Trie t: node.next) {
                 if (patternSearch(word, t, index + 1)) return true;
             }
         } else {
            if (patternSearch(word, node.next[c], index + 1)) return true;
         }        
         return false;
    }
    
    class Trie {
        Trie[] next = new Trie[R];
        boolean isWord;
    }    
}

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

Solution 2021-10-23

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class WordDictionary {

    TreeNode root = new TreeNode();
    
    public WordDictionary() {
        
    }
    
    /*
    bad
    
    *
    [b]
      [a]
    
    
    */
    public void addWord(String word) {
        TreeNode node = root;
        for (char c: word.toCharArray()) {
            if (node.nodes[c - 'a']  == null) {
                node.nodes[c - 'a'] = new TreeNode();
            }
            node = node.nodes[c - 'a'];
        }
        node.isTerminal = true;
    }
    
    public boolean search(String word) {
         return search(root, 0, word);
    }
    
    boolean search(TreeNode node, int index, String word) {
         if (node == null) {
             return false;
         }   
        
         if (index == word.length()) {
             return node.isTerminal;
         }
         
         char c = word.charAt(index);
         if (c == '.') {
            for (TreeNode searchNode: node.nodes) {
                boolean find = search(searchNode, index + 1, word);
                if (find) {
                    return true;
                }
            }
         } else {
            return search(node.nodes[c - 'a'], index + 1, word);
         }
        return false;
    }
    
    class TreeNode {
        TreeNode[] nodes = new TreeNode[26];
        boolean isTerminal;
    }
}

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

Solution 2021-01-30

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class WordDictionary {
    
    Node root = new Node();

    public WordDictionary() {
        
    }
    
    public void addWord(String word) {
        Node curr = root;
        for (char c: word.toCharArray()) {
            if (curr.child[c] == null) {
                curr.child[c] = new Node();
            }
            curr = curr.child[c];
        }
        curr.isTerminal = true;
    }
    
    public boolean search(String word) {
         return search(root, 0, word);
    }
    
    boolean search(Node curr, int index, String word) {
        if (curr == null) return false;
        if (index == word.length()) {
            return curr.isTerminal;
        }

        char c = word.charAt(index);
        
        if (c == '.') {
            for (Node n: curr.child) {
                if (search(n, index + 1, word)) {
                    return true;
                }
            }
            return false;
        } else {
            return search(curr.child[c], index + 1, word);
        }
    }
    
    
    class Node {
        Node[] child = new Node[256];
        boolean isTerminal;
    }
}

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