algorithms/algorithms/backtrack/pattern_match.py at master · pk210700/algorithms · GitHub

Repository navigation

42 lines (34 loc) · 1.29 KB

"""

Given a pattern and a string str,

find if str 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 substring in str.

Examples:

pattern = "abab", str = "redblueredblue" should return true.

pattern = "aaaa", str = "asdasdasdasd" should return true.

pattern = "aabb", str = "xyzabcxzyabc" should return false.

Notes:

You may assume both pattern and str contains only lowercase letters.

"""

def pattern_match(pattern, string):

"""

:type pattern: str

:type string: str

:rtype: bool

"""

def backtrack(pattern, string, dic):

if len(pattern) == 0 and len(string) > 0:

return False

if len(pattern) == len(string) == 0:

return True

for end in range(1, len(string)-len(pattern)+2):

if pattern[0] not in dic and string[:end] not in dic.values():

dic[pattern[0]] = string[:end]

if backtrack(pattern[1:], string[end:], dic):

return True

del dic[pattern[0]]

elif pattern[0] in dic and dic[pattern[0]] == string[:end]:

if backtrack(pattern[1:], string[end:], dic):

return True

return False

return backtrack(pattern, string, {})