LeetCode: Implement Magic Dictionary – Prepare For Coder Interview

Implement Magic Dictionary

Similar Problems:

Implement a magic directory with buildDict, and search methods.

For the method buildDict, you’ll be given a list of non-repetitive words to build a dictionary.

For the method search, you’ll be given a word, and judge whether if you modify exactly one character into another character in this word, the modified word is in the dictionary you just built.

Example 1:

Input: buildDict(["hello", "leetcode"]), Output: Null
Input: search("hello"), Output: False
Input: search("hhllo"), Output: True
Input: search("hell"), Output: False
Input: search("leetcoded"), Output: False

Note:

You may assume that all the inputs are consist of lowercase letters a-z.
For contest purpose, the test data is rather small by now. You could think about highly efficient algorithm after the contest.
Please remember to RESET your class variables declared in class MagicDictionary, as static/class variables are persisted across multiple test cases. Please see here for more details.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

Solution: Trie + recursive

// https://code.dennyzhang.com/implement-magic-dictionary
// Basic Ideas: Trie tree
//
//  Strict match and fuzz match
//
//  The time complexity would be 26*n
//
// Complexity: Time O(n), Space O(n)
type MagicDictionary struct {
    children map[byte]*MagicDictionary
    isLeaf bool
}

/** Initialize your data structure here. */
func Constructor() MagicDictionary {
    return MagicDictionary{children:map[byte]*MagicDictionary{}}
}

/** Build a dictionary through a list of words */
func (this *MagicDictionary) BuildDict(dict []string)  {
    for _, w := range dict {
        p := this
        for i, _ := range w {
            ch := w[i]
            if _, ok := p.children[ch]; !ok {
                q := &MagicDictionary{children:map[byte]*MagicDictionary{}}
                p.children[ch] = q
            }
            p = p.children[ch]
        }
        p.isLeaf = true
    }
}

func (this *MagicDictionary) strictSearch(word string) bool {
    p := this
    for i, _ := range word {
        ch := word[i]
        if _, ok := p.children[ch]; !ok {
            return false
        }
        p = p.children[ch]
    }
    return p.isLeaf
}

/** Returns if there is any word in the trie that equals to the given word after modifying exactly one character */
func (this *MagicDictionary) Search(word string) bool {
    if len(word) == 0 {
        return false
    }
    for b, q := range this.children {
        ch := word[0]
        if b != ch {
            if q.strictSearch(word[1:]) {
                return true
            }
        } else {
            if q.Search(word[1:]) {
                return true
            }
        }
    }
    return false
}

/**
 * Your MagicDictionary object will be instantiated and called as such:
 * obj := Constructor();
 * obj.BuildDict(dict);
 * param_2 := obj.Search(word);
 */

Solution: Trie

## https://code.dennyzhang.com/implement-magic-dictionary
## Basic Ideas: Trie Tree + recursive
##
## Complexity: Time O(w*h): w is the width, h is the height
##             Space O(n): n sum of all characters in dict
import collections
class TrieNode:
    def __init__(self):
        self.children = collections.defaultdict(TrieNode)
        self.is_word = False

class Trie:
    def __init__(self):
        self.root = TrieNode()

    def addWord(self, trieNode, word):
        node = trieNode
        for ch in word:
            node = node.children[ch]
        node.is_word = True

    def searchWord(self, trieNode, word):
        node = trieNode
        for ch in word:
            if ch not in node.children:
                return False
            node = node.children[ch]
        return node.is_word

class MagicDictionary:

    def __init__(self):
        """
        Initialize your data structure here.
        """
        self.myTrie = Trie()


    def buildDict(self, dict):
        """
        Build a dictionary through a list of words
        :type dict: List[str]
        :rtype: void
        """
        for word in dict:
            self.myTrie.addWord(self.myTrie.root, word)

    def search(self, word):
        """
        Returns if there is any word in the trie that equals to the given word after modifying exactly one character
        :type word: str
        :rtype: bool
        """
        if len(word) == 0: return False
        (status, cnt) = self.mySearch(self.myTrie, self.myTrie.root, word, False)
        return (status is True) and (cnt == 1)

    def mySearch(self, trie, trieNode, word, exactMatch):
        """
        :rtype: (bool, cnt)
        """
        # base cases for recursive
        if exactMatch == True:
            status = trie.searchWord(trieNode, word)
            return (status, 0)

        if len(word) == 1:
            for ch in trieNode.children:
                if word[0] != ch and trieNode.children[ch].is_word:
                    return (True, 1)
            return (False, 0)

        for ch in trieNode.children:
            # difference happens in the next layer
            if word[0] == ch:
                (status, cnt) = self.mySearch(trie, trieNode.children[ch], word[1:], False)
                if (status is True) and (cnt == 1): return (True, 1)
            else:
                # difference happens in the current layer
                (status, cnt) = self.mySearch(trie, trieNode.children[ch], word[1:], True)
                if (status is True) and (cnt == 0): return (True, 1)
        return (False, 0)

# Your MagicDictionary object will be instantiated and called as such:
# obj = MagicDictionary()
# obj.buildDict(dict)
# param_2 = obj.search(word)

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