Medium
Design Memory Allocator — Python
Full explanation · Time ctor: O(1) allocate: O(logn) free: O(logn) · Space O(n)
# Time: ctor: O(1)
# allocate: O(logn)
# free: O(logn)
# Space: O(n)
from sortedcontainers import SortedList
import collections
# sorted list
class Allocator(object):
def __init__(self, n):
"""
:type n: int
"""
self.__avails = SortedList([[0, n]])
self.__lookup = collections.defaultdict(list)
def allocate(self, size, mID):
"""
:type size: int
:type mID: int
:rtype: int
"""
for l, s in self.__avails:
if s < size:
continue
self.__avails.remove([l, s])
self.__lookup[mID].append([l, size])
if s-size > 0:
self.__avails.add([l+size, s-size])
return l
return -1
def free(self, mID):
"""
:type mID: int
:rtype: int
"""
if mID not in self.__lookup:
return 0
result = 0
for l, s in self.__lookup[mID]:
self.__avails.add([l, s])
i = self.__avails.bisect_left([l, s])
if i+1 < len(self.__avails) and self.__avails[i][0]+self.__avails[i][1] == self.__avails[i+1][0]:
self.__avails[i][1] += self.__avails[i+1][1]
del self.__avails[i+1]
if i-1 >= 0 and self.__avails[i-1][0]+self.__avails[i-1][1] == self.__avails[i][0]:
self.__avails[i-1][1] += self.__avails[i][1]
del self.__avails[i]
result += s
del self.__lookup[mID]
return result