Word Pattern

Problem

https://leetcode.com/problems/word-pattern/

Given a pattern and a string s, find if s follows the same pattern.

Here follow means a full match, such that there is a bijection between a letter in pattern and a non-empty word in s. Specifically:

Each letter in pattern maps to exactly one unique word in s.
Each unique word in s maps to exactly one letter in pattern.
No two letters map to the same word, and no two words map to the same letter.

Example 1:

Input: pattern = “abba”, s = “dog cat cat dog”

Output: true

Explanation:

The bijection can be established as:

'a' maps to "dog".
'b' maps to "cat".

Example 2:

Input: pattern = “abba”, s = “dog cat cat fish”

Output: false

Example 3:

Input: pattern = “aaaa”, s = “dog cat cat dog”

Output: false

Constraints:

1 <= pattern.length <= 300
pattern contains only lower-case English letters.
1 <= s.length <= 3000
s contains only lowercase English letters and spaces ' '.
s does not contain any leading or trailing spaces.
All the words in s are separated by a single space.

Pattern

Hash Table, String

Approaches

Hash Map

Explanation

Split s into a list of words. The problem reduces to IsomorphicStrings but with characters in pattern and elements in s. Since pattern and s can have different lenghts, we first check that they have the same length.

Code

def wordPattern(pattern: str, s: str) -> bool:
    """Given a pattern and a string ``s``, finds if ``s`` follows the same pattern."""
    words = s.split()

    if len(pattern) != len(words):
        return False

    d = {}
    d_range = set()
    for char, word in zip(pattern, words):
        if char in d:
            if d[char] != word:
                return False
        else:
            if word in d_range:
                return False
            else:
                d[char] = word
                d_range.add(word)

    return True

Test

>>> from word_pattern__hash_map import wordPattern
>>> wordPattern('abba', 'dog cat cat dog')
True
>>> wordPattern('abba', 'dog cat cat fish')
False
>>> wordPattern('aaaa', 'dog cat cat dog')
False

Complexity

\(n\) is the length of pattern and the number of words in s

Measure	Complexity	Notes
Time	\(O(n)\)	single pass over `pattern` and words in `s`
Auxiliary Space	\(O(n)\)	dictionary for mapping characters to words and set for tracking mapped words

word_pattern__hash_map.wordPattern(pattern: str, s: str) → bool: Given a pattern and a string s, finds if s follows the same pattern.