Sponsored
Sponsored
In this approach, we establish a one-to-one mapping for each letter in the pattern to letters in each word. To do this, we use two maps or dictionaries: one for mapping pattern to word letters and another for mapping word to pattern letters. Both maps are used to ensure that the mapping is bijective (i.e., one-to-one).
Time Complexity: O(n * m), where n is the number of words and m is the word length.
Space Complexity: O(m), for the maps storing character mappings.
1def matchesPattern(word, pattern):
2 mapPtoW, mapWtoP = {}, {}
3 for w, p in zip(word, pattern):
4 if p not in mapPtoW and w not in mapWtoP:
5 mapPtoW[p] = w
6 mapWtoP[w] = p
7 elif mapPtoW.get(p) != w or mapWtoP.get(w) != p:
8 return False
9 return True
10
11
12def findAndReplacePattern(words, pattern):
13 return [word for word in words if matchesPattern(word, pattern)]
14This Python code leverages dictionaries to store character mappings from pattern to word and vice versa. The function matchesPattern is a helper used within findAndReplacePattern to filter out non-matching words.
In this approach, we map each character of the string to its first occurrence index in the string. This creates a unique encoding pattern for comparison. Two strings match if they have the same encoding pattern.
Time Complexity: O(n * m), where n is the number of words and m is the word length.
Space Complexity: O(m), for storing encoded strings and maps.
1
This C program calculates an encoded representation for the pattern and each word using indices of first occurrences. It then compares these encodings to determine if a word matches the pattern.