# 3161. Block Placement Queries
from sortedcontainers import SortedList
from itertools import pairwise
from bisect import bisect_right
"""
there exists an infinite number line, with its origin at 0 and extending
towards the positive x-axis. you are given a 2d array 'qeuries', which
contains two types of queries: for a query of type 1, 'queries{i} = {1, x}'.
build an obstacle at distance 'x' from the origin. it is guaranteed that
there is no obstacle at distance 'x' when the query is asked. for query type
2, 'queries{i} = {2, x, sz}'. check if it is possible to place a block of
size 'sz' anywhere in the range '{0, x}' on the line, such that the block
entirely lies in the range '{0, x}'. a block cannot be placed if it
intersects with any obstacle, but it may touch it. note that you do not
actually place the block. queries are separate. return a boolean array
'results' where 'results{i}' is true if you can place a block specified in
the i'th query of type 2, and false otherwise.
"""
class tree:
def __init__(self, n):
self.n = n
self.l = [0] * (n + 1)
def add(self, i, x):
i += 1
while i <= self.n:
self.l[i] = max(self.l[i], x)
i += i & -i
def query(self, i):
i += 1
ans = 0
while i:
ans = max(ans, self.l[i])
i -= i & -i
return ans
class Solution(object):
def getResults(self, queries):
"""
:type queries: List[List[int]]
:rtype: List[bool]
"""
list = SortedList()
n = min(5 * 10**4, len(queries) * 3)
list.add(0)
list.add(n)
for i in queries:
if i[0] == 1:
curr = i[1]
list.add(curr)
bit, ans = tree(n + 1), []
for i, j in pairwise(list):
bit.add(j, j - i)
for i in reversed(queries):
if i[0] == 1:
curr = i[1]
idx = list.index(curr)
post = list[idx + 1]
prev = list[idx - 1]
list.remove(curr)
bit.add(post, post - prev)
else:
_, curr, size = i
idx = list.bisect_right(curr)
prev = list[idx - 1]
ans.append(bit.query(prev) >= size or (curr - prev) >= size)
ans.reverse()
return ans
if __name__ == "__main__":
obj = Solution()
print(obj.getResults(queries=[[1, 2], [2, 3, 3], [2, 3, 1], [2, 2, 2]]))
print(obj.getResults(queries=[[1, 7], [2, 7, 6], [1, 2], [2, 7, 5], [2, 7, 6]]))
