Hard
Number of Subarrays That Match a Pattern II — Python
Full explanation · Time O(n) · Space O(m)
# Time: O(n)
# Space: O(m)
# kmp
class Solution(object):
def countMatchingSubarrays(self, nums, pattern):
"""
:type nums: List[int]
:type pattern: List[int]
:rtype: int
"""
def getPrefix(pattern):
prefix = [-1]*len(pattern)
j = -1
for i in xrange(1, len(pattern)):
while j+1 > 0 and pattern[j+1] != pattern[i]:
j = prefix[j]
if pattern[j+1] == pattern[i]:
j += 1
prefix[i] = j
return prefix
def KMP(text, pattern):
prefix = getPrefix(pattern)
j = -1
for i, x in enumerate(text):
while j+1 > 0 and pattern[j+1] != x:
j = prefix[j]
if pattern[j+1] == x:
j += 1
if j+1 == len(pattern):
yield i-j
j = prefix[j]
return sum(1 for _ in KMP((cmp(nums[i+1], nums[i]) for i in xrange(len(nums)-1)), pattern))