LeetCode Tutorial in Python 290. Word Pattern

Published: 13 September 2023
on channel: NeedCode

LeetCode Tutorial in Python 290. Word Pattern #python #leetcode

The intuition behind this solution is to determine if there is a bijection (a one-to-one correspondence) between the characters in the given pattern and the words in the input string s.

Here's a step-by-step breakdown of the intuition:

Splitting the String: The input string s is split into individual words using the s.split() method. This step helps separate the words in the string, making it easier to compare them to the characters in the pattern.

Checking Unique Characters and Words:

len(set(pattern)): This part calculates the number of unique characters in the pattern. For example, if the pattern is "abba," there are only two unique characters, 'a' and 'b'.

len(set(s)): This calculates the number of unique words in the string s. Each word in the string is counted only once.

The goal here is to ensure that the number of unique characters in the pattern matches the number of unique words in the string. This is important because for a bijection, there should be a one-to-one mapping between characters and words.

Checking Bijection with zip:

set(zip(pattern, s)): This part pairs each character in the pattern with its corresponding word from the split string s. For example, if pattern = "abba" and s = "dog cat cat dog", it would create the pairs: ('a', 'dog'), ('b', 'cat'), ('b', 'cat'), and ('a', 'dog').

The set function is used to ensure that there are no duplicate pairs. The goal here is to check if there is a one-to-one mapping between characters and words. If there are any duplicate pairs, it indicates that the bijection condition is not met.

Final Check:

len(pattern) == len(s): This condition ensures that the lengths of the pattern and the string s are equal. This is necessary for a bijection to exist because there must be a one-to-one correspondence between characters and words.

Overall Comparison: All of these conditions are combined using and to create a single boolean expression. If all conditions are met, the function returns True, indicating that the bijection exists and the input string s follows the same pattern as the given pattern.

In summary, the intuition is to break down the problem into smaller steps:

Check the number of unique characters in the pattern and the number of unique words in the string.

Pair each character with its corresponding word.

Ensure there are no duplicate pairs and that the lengths of the pattern and string match.

If all these conditions are met, the function returns True, indicating that the string follows the same pattern as specified. Otherwise, it returns False.

Time Complexity:

Calculating the length of a set based on a string or a pattern has a time complexity of O(n), where n is the length of the string or pattern.

Creating the zip object using zip(pattern, s) also has a time complexity of O(n), where n is the length of the shorter input (either pattern or s).

In the solution, we perform three set operations (one for the pattern, one for the split string, and one for the zip result), all of which are O(n).

Therefore, the overall time complexity is O(n).

Space Complexity:

Creating sets based on the pattern and the split string s requires additional space. In the worst case, when all characters or words are unique, the space complexity for creating these sets is O(n).

Creating the zip object and then converting it to a set also requires additional space. In the worst case, when all characters in the pattern are distinct from all words in s, this would also be O(n).

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