Description

Given an integer array bloomDay, an integer m and an integer k.

We need to make m bouquets. To make a bouquet, you need to use k adjacent flowers from the garden.

The garden consists of n flowers, the ith flower will bloom in the bloomDay[i] and then can be used in exactly one bouquet.

Return the minimum number of days you need to wait to be able to make m bouquets from the garden. If it is impossible to make m bouquets return -1.

Example 1:

Input: bloomDay = [1,10,3,10,2], m = 3, k = 1
Output: 3
Explanation: Let's see what happened in the first three days. x means flower bloomed and _ means flower didn't bloom in the garden.
We need 3 bouquets each should contain 1 flower.
After day 1: [x, _, _, _, _]   // we can only make one bouquet.
After day 2: [x, _, _, _, x]   // we can only make two bouquets.
After day 3: [x, _, x, _, x]   // we can make 3 bouquets. The answer is 3.

Example 2:

Input: bloomDay = [1,10,3,10,2], m = 3, k = 2
Output: -1
Explanation: We need 3 bouquets each has 2 flowers, that means we need 6 flowers. We only have 5 flowers so it is impossible to get the needed bouquets and we return -1.

Example 3:

Input: bloomDay = [7,7,7,7,12,7,7], m = 2, k = 3
Output: 12
Explanation: We need 2 bouquets each should have 3 flowers.
Here's the garden after the 7 and 12 days:
After day 7: [x, x, x, x, _, x, x]
We can make one bouquet of the first three flowers that bloomed. We cannot make another bouquet from the last three flowers that bloomed because they are not adjacent.
After day 12: [x, x, x, x, x, x, x]
It is obvious that we can make two bouquets in different ways.

Example 4:

Input: bloomDay = [1000000000,1000000000], m = 1, k = 1
Output: 1000000000
Explanation: You need to wait 1000000000 days to have a flower ready for a bouquet.

Example 5:

Input: bloomDay = [1,10,2,9,3,8,4,7,5,6], m = 4, k = 2
Output: 9

Constraints:

  • bloomDay.length == n
  • 1 <= n <= 10^5
  • 1 <= bloomDay[i] <= 10^9
  • 1 <= m <= 10^6
  • 1 <= k <= n

分析

题目的意思是：给你一个数组，表示的是花开放的时间，要求开m次花，每次有K朵相邻的花盛开。求出所用的最小时间。题目很拗口，我也没看懂，这是一个二叉查找树的题目，首先要在数组中最小值和最大值之间找到一个满足条件的数，然后找出刚好满足条件的数，需要用到二叉查找来缩小查找范围。如果能够想到这个就能做出来了。在二叉查找的时候判断是否满足条件就行了，用low和high表示二叉查找的边界，如果判断过大，则high=mid-1，如果过小则mid+1。终止条件当然是low>high啊，返回low就是第一个满足条件的值了。

代码

class Solution:
    def bloomCount(self,bloomDay,t,k):
        n=len(bloomDay)
        count=0
        res=0
        for i in range(n):
            if(bloomDay[i]<=t):
                count+=1
                if(count==k):
                    count=0
                    res+=1
            else:
                count=0
        return res
        
    def minDays(self, bloomDay: List[int], m: int, k: int) -> int:
        low,high=min(bloomDay),max(bloomDay)
        if(m*k>len(bloomDay)):
            return -1
        while(low<=high):
            mid=low+(high-low)//2
            if(self.bloomCount(bloomDay,mid,k)<m):
                low=mid+1
            else:
                high=mid-1
        return low

参考文献

​​[LeetCode] [Python] Binary Search Solution​​