Medium
Fruits Into Baskets III — Python
Full explanation · Time O(nlogn) · Space O(n)
# Time: O(nlogn)
# Space: O(n)
# segment tree, binary search
class Solution(object):
def numOfUnplacedFruits(self, fruits, baskets):
"""
:type fruits: List[int]
:type baskets: List[int]
:rtype: int
"""
class SegmentTree(object):
def __init__(self, N,
build_fn=lambda _: 0,
query_fn=lambda x, y: y if x is None else x if y is None else max(x, y),
update_fn=lambda x: x):
self.tree = [None]*(2*2**((N-1).bit_length()))
self.base = len(self.tree)//2
self.query_fn = query_fn
self.update_fn = update_fn
for i in xrange(self.base, self.base+N):
self.tree[i] = build_fn(i-self.base)
for i in reversed(xrange(1, self.base)):
self.tree[i] = query_fn(self.tree[2*i], self.tree[2*i+1])
def update(self, i, h):
x = self.base+i
self.tree[x] = self.update_fn(h)
while x > 1:
x //= 2
self.tree[x] = self.query_fn(self.tree[x*2], self.tree[x*2+1])
def binary_search(self, x):
if self.tree[1] < x:
return -1
i = 1
while not i >= self.base:
if self.tree[2*i] >= x:
i = 2*i
else:
i = 2*i+1
return i-self.base
def build(i):
return baskets[i]
st = SegmentTree(len(baskets), build_fn=build)
result = 0
for x in fruits:
i = st.binary_search(x)
if i == -1:
result += 1
else:
st.update(i, 0)
return result