Naive Solution

The naive solution involves iterating through the dictionary for each search word and comparing each word in the dictionary with the search word. For each comparison, we count the number of differing characters. If exactly one character differs, we return true. Otherwise, we continue the search. If we reach the end of the dictionary without finding a match, we return false.

Code (Python)

class MagicDictionary:
    def __init__(self):
        self.dictionary = []

    def buildDict(self, dictionary: List[str]) -> None:
        self.dictionary = dictionary

    def search(self, searchWord: str) -> bool:
        for word in self.dictionary:
            if len(word) == len(searchWord):
                diff_count = 0
                for i in range(len(word)):
                    if word[i] != searchWord[i]:
                        diff_count += 1
                if diff_count == 1:
                    return True
        return False

Time Complexity

• buildDict: O(N), where N is the number of words in the input dictionary.
• search: O(M * L), where M is the number of words in the dictionary and L is the length of the search word.

Space Complexity

• O(N), where N is the number of words in the dictionary, to store the dictionary.

Optimal Solution

A slightly more optimal solution could involve using a hash map to store words grouped by their length. During the search, only words with the same length as the searchWord need to be considered. This approach reduces the number of unnecessary comparisons.

Code (Python)

class MagicDictionary:
    def __init__(self):
        self.length_map = {}

    def buildDict(self, dictionary: List[str]) -> None:
        for word in dictionary:
            length = len(word)
            if length not in self.length_map:
                self.length_map[length] = []
            self.length_map[length].append(word)

    def search(self, searchWord: str) -> bool:
        length = len(searchWord)
        if length not in self.length_map:
            return False

        for word in self.length_map[length]:
            diff_count = 0
            for i in range(length):
                if word[i] != searchWord[i]:
                    diff_count += 1
            if diff_count == 1:
                return True
        return False

Time Complexity

• buildDict: O(N), where N is the number of words in the input dictionary.
• search: O(K * L), where K is the number of words with the same length as the search word, and L is the length of the search word. In the worst case, K can be equal to N, so it can still be O(N*L), but on average it's better if the word lengths are varied.

Space Complexity

• O(N), where N is the number of words in the dictionary, to store the length_map.

Edge Cases

1. Empty dictionary.
2. Empty search word (though constraints specify length >= 1).
3. No words in the dictionary with the same length as the search word.

These edge cases are handled correctly in the provided implementations.