How do you solve Words Within Two Edits of Dictionary on LeetCode?

Words Within Two Edits of Dictionary (LeetCode #2452) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * k * 26^2) time complexity and O(n) space complexity. Key insight: Understanding character differences is key to solving this problem.

What is the brute force solution for Words Within Two Edits of Dictionary?

The brute force approach for Words Within Two Edits of Dictionary is Brute Force, with O(n²) time complexity and O(1) space complexity. We will compare each word in the queries with every word in the dictionary. For each pair, we will count the number of differing characters. If the count is 2 or less, we consider it a match. The optimal Optimal Solution approach improves this to O(n * k * 26^2).

What algorithm or data structure is used to solve Words Within Two Edits of Dictionary?

Words Within Two Edits of Dictionary uses the following concepts: Array, String, Trie. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Words Within Two Edits of Dictionary?

Clarify the problem requirements before jumping into coding. Think about edge cases, such as words that are already in the dictionary. Practice generating combinations or variations of strings.

What are common mistakes when solving Words Within Two Edits of Dictionary?

Not considering all possible edits, especially when generating new words. Overlooking the need to break early once a match is found.

#2452

Words Within Two Edits of Dictionary

Medium

ArrayStringTrieHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * k * 26^2)
Space	O(1)	O(n)

💡

Intuition

Time O(n * k * 26^2)Space O(n)

Instead of comparing each query with every dictionary word, we can use a set to efficiently check for potential matches by generating all possible words that are two edits away from each query.

⚙️

Algorithm

5 steps

1Step 1: Create a set to store all words in the dictionary for O(1) lookups.
2Step 2: For each word in queries, generate all possible words with 0, 1, or 2 edits.
3Step 3: Check if any of the generated words exist in the dictionary set.
4Step 4: If a match is found, add the query word to the results list.
5Step 5: Return the results list.

solution.py20 lines

1def generate_words(word):
2    alphabet = 'abcdefghijklmnopqrstuvwxyz'
3    words = set()
4    for i in range(len(word)):
5        for c in alphabet:
6            if c != word[i]:
7                words.add(word[:i] + c + word[i+1:])
8                for c2 in alphabet:
9                    if c2 != word[i] and c2 != c:
10                        words.add(word[:i] + c2 + word[i+1:])
11    return words
12
13def two_edit_words(queries, dictionary):
14    dict_set = set(dictionary)
15    result = []
16    for query in queries:
17        possible_words = generate_words(query)
18        if any(word in dict_set for word in possible_words):
19            result.append(query)
20    return result

ℹ

Complexity note: This complexity arises from generating potential words for each query with two edits, where k is the length of the words (constant). The space complexity is O(n) due to the dictionary set.

1Understanding character differences is key to solving this problem.
2Using a set for quick lookups can significantly improve efficiency.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.