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。


  • 返回:

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


利用 Fenwick tree ft[k]存储前k个槽一共有多少朵花,则区间[m, n]的花朵总数 = ft[n] - ft[m - 1]



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: \(O(N)\).


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

同样使用Fenwick tree, 但是要注意检测是否前一天满足条件的group存在, 并且当天该组两侧没有花开, 如果当天开得花破坏了已有group, 再寻找是否存在新的group满足条件。

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)
        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