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maximum-length-of-repeated-subarray.py 3.01 KB
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Allen Liu 提交于 7年前 . add complexity
# Time: O(m * n)
# Space: O(min(m, n))
import collections
class Solution(object):
def findLength(self, A, B):
"""
:type A: List[int]
:type B: List[int]
:rtype: int
"""
if len(A) < len(B): return self.findLength(B, A)
result = 0
dp = [[0] * (len(B)+1) for _ in xrange(2)]
for i in xrange(len(A)):
for j in xrange(len(B)):
if A[i] == B[j]:
dp[(i+1)%2][j+1] = dp[i%2][j]+1
else:
dp[(i+1)%2][j+1] = 0
result = max(result, max(dp[(i+1)%2]))
return result
# Time: O(m * n * log(min(m, n)))
# Space: O(min(m, n))
# Binary search + rolling hash solution (226 ms)
class Solution2(object):
def findLength(self, A, B):
"""
:type A: List[int]
:type B: List[int]
:rtype: int
"""
if len(A) > len(B): return self.findLength(B, A)
M, p = 10**9+7, 113
p_inv = pow(p, M-2, M)
def check(guess):
def rolling_hashes(source, length):
if length == 0:
yield 0, 0
return
val, power = 0, 1
for i, x in enumerate(source):
val = (val + x*power) % M
if i < length - 1:
power = (power*p) % M
else:
yield val, i-(length-1)
val = (val-source[i-(length-1)])*p_inv % M
hashes = collections.defaultdict(list)
for hash_val, i in rolling_hashes(A, guess):
hashes[hash_val].append(i)
for hash_val, j in rolling_hashes(B, guess):
if any(A[i:i+guess] == B[j:j+guess] for i in hashes[hash_val]):
return True
return False
left, right = 0, min(len(A), len(B)) + 1
while left < right:
mid = left + (right-left)/2
if not check(mid): # find the min idx such that check(idx) == false
right = mid
else:
left = mid+1
return left-1
# Time: O(m * n * min(m, n) * log(min(m, n)))
# Space: O(min(m^2, n^2))
# Binary search (122 ms)
class Solution3(object):
def findLength(self, A, B):
"""
:type A: List[int]
:type B: List[int]
:rtype: int
"""
if len(A) > len(B): return self.findLength(B, A)
def check(length):
lookup = set(A[i:i+length] \
for i in xrange(len(A)-length+1))
return any(B[j:j+length] in lookup \
for j in xrange(len(B)-length+1))
A = ''.join(map(chr, A))
B = ''.join(map(chr, B))
left, right = 0, min(len(A), len(B)) + 1
while left < right:
mid = left + (right-left)/2
if not check(mid): # find the min idx such that check(idx) == false
right = mid
else:
left = mid+1
return left-1
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