How do you solve Find and Replace Pattern on LeetCode?

Find and Replace Pattern (LeetCode #890) can be solved using Brute Force. The optimal approach is Brute Force with undefined time complexity and undefined space complexity. Key insight: We can check each word against the pattern by creating a mapping of characters in the pattern to characters in the word. If we can establish a one-to-one mapping for all characters, the word matches the pattern.

What is the time complexity of Find and Replace Pattern?

The optimal solution for Find and Replace Pattern runs in undefined time and uses undefined space. undefined

What algorithm or data structure is used to solve Find and Replace Pattern?

Find and Replace Pattern uses the following concepts: Array, Hash Table, String. The recommended patterns to study are: .

#890

Find and Replace Pattern

Medium

ArrayHash TableString

LeetCode ↗

Approaches

💡

Intuition

Time Space

We can check each word against the pattern by creating a mapping of characters in the pattern to characters in the word. If we can establish a one-to-one mapping for all characters, the word matches the pattern.

⚙️

Algorithm

5 steps

1Step 1: For each word in the list, check if its length matches the pattern's length.
2Step 2: Create two dictionaries: one to map characters from the pattern to the word and another to map characters from the word to the pattern.
3Step 3: Iterate through the characters of the pattern and the word simultaneously, updating the dictionaries.
4Step 4: If a character mapping is inconsistent (i.e., a character in the pattern maps to different characters in the word), the word does not match.
5Step 5: If all characters can be mapped consistently, add the word to the result list.

solution.py13 lines

1def findAndReplacePattern(words, pattern):
2    def matches(word):
3        if len(word) != len(pattern): return False
4        p_to_w, w_to_p = {}, {}
5        for p_char, w_char in zip(pattern, word):
6            if p_char not in p_to_w:
7                p_to_w[p_char] = w_char
8            if w_char not in w_to_p:
9                w_to_p[w_char] = p_char
10            if p_to_w[p_char] != w_char or w_to_p[w_char] != p_char:
11                return False
12        return True
13    return [word for word in words if matches(word)]

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