Hard
Count Integers in Intervals — Python
Full explanation · Time ctor: O(1) add: O(logn), amortized Count: O(1) · Space O(n)
# Time: ctor: O(1)
# add: O(logn), amortized
# count: O(1)
# Space: O(n)
from sortedcontainers import SortedList
# design, sortedlist
class CountIntervals(object):
def __init__(self):
self.__sl = SortedList()
self.__cnt = 0
def add(self, left, right):
"""
:type left: int
:type right: int
:rtype: None
"""
i = self.__sl.bisect_right((left,))
if i-1 >= 0 and self.__sl[i-1][1]+1 >= left:
i -= 1
left = self.__sl[i][0]
to_remove = []
for i in xrange(i, len(self.__sl)):
if not (right+1 >= self.__sl[i][0]):
break
right = max(right, self.__sl[i][1])
self.__cnt -= self.__sl[i][1]-self.__sl[i][0]+1
to_remove.append(i)
while to_remove:
del self.__sl[to_remove.pop()]
self.__sl.add((left, right))
self.__cnt += right-left+1
def count(self):
"""
:rtype: int
"""
return self.__cnt