# 683 K Empty Slots

Original Description
There is a garden with N slots. In each slot, there is a flower. The N flowers will bloom one by one in N days. In each day, there will be exactly one flower blooming and it will be in the status of blooming since then.
Given an array flowers consists of number from 1 to N. Each number in the array represents the place where the flower will open in that day.
For example, flowers[i] = x means that the unique flower that blooms at day i will be at position x, where i and x will be in the range from 1 to N.
Also given an integer k, you need to output in which day there exists two flowers in the status of blooming, and also the number of flowers between them is k and these flowers are not blooming.
If there isn’t such day, output -1.

• 要求： 花园中有N个槽，给定种花的顺序flowers，flowers[i] = x 表示第i天，在第x个槽种下一朵花。

给定数字k，求flowers中是否存在某一天，满足相隔k距离的两个端点恰好各有一朵花，而这两朵花之间的k个槽都没有花。

• 返回：
若有一天符合要求，返回该时间，否则返回-1

• 例子：
• input:
flowers: [1,3,2], k=1
• output: 2 (In the second day, the first and the third flower have become blooming.)

## Solution

Reference

class FenwickTree(object):
def __init__(self, n):
self.n = n
self.tree = [0] * (n + 1)
def lowbit(self, x):
return x & -x
def update(self, x, val):  # update value at each index
while x <= self.n:
self.tree[x] += val
x += self.lowbit(x)
def sum(self, x):         # return sum of first x elements
res = 0
while x > 0:
res += self.tree[x]
x -= self.lowbit(x)
return res
class Solution(object):
def kEmptySlots(self, flowers, k):
"""
:type flowers: List[int]
:type k: int
:rtype: int
"""
maxn = max(flowers)
nums = [0] * (maxn + 1)
ft = FenwickTree(maxn)
for day_index, position in enumerate(flowers):
ft.update(position, 1)  # update value
nums[position] = 1      # this position has a flower open on the day
if position >= k and ft.sum(position) - ft.sum(position - k - 2) == 2 and nums[position - k - 1]:
return day_index + 1  # as index starts from 0
if position + k + 1<= maxn and ft.sum(position + k + 1) - ft.sum(position - 1) == 2 and nums[position + k + 1]:
return day_index + 1
return -1


Time complexity:$$O(NlogN)$$ (as sum/update operation in fenwick tree is O(logN)), space complexity: $$O(N)$$.

## Variation:

output the lastest day that contains a group of K flowers.

class Solution(object):
def Kgroup(self, P, K):
N = len(P)
ftree = FenwickTree(N)
nums = [0] * (N+2)
day = -1
old_position = 0
for day_index, position in enumerate(P):
ftree.update(position, 1)
nums[position] = 1
# check if group in the previous day still exist
if day!=-1 and self.checkgroup(K, ftree, nums, P[old_position]) == True:
day = day_index+1
# otherwise search for new group
elif self.checkgroup(K, ftree, nums, position) == True:
day = day_index+1
old_position = position
print(day, nums)
print(day)
return day
def checkgroup(self, K, ftree, nums, position):
if position >= K and ftree.sum(position)- ftree.sum(position-K) == K and nums[position-K] ==0 and nums[position+1]==0:
return True
if position + K <= len(nums)-2 and ftree.sum(position+K)- ftree.sum(position)==K and nums[position+K] == 0 and nums[position-1]==0:
return True