<span>leetcode-1375</span>

There is a room with n bulbs, numbered from 1 to n, arranged in a row from left to right. Initially, all the bulbs are turned off.At moment k (for k from 0 to n - 1), we turn on the light[k] bulb. A bulb change color to blue only if it is on and all the previous bulbs (to the left) are turned on too.

Return the number of moments in which all turned on bulbs are blue.

Input: light = [2,1,3,5,4]
Output: 3
Explanation: All bulbs turned on, are blue at the moment 1, 2 and 4.

  

Input: light = [3,2,4,1,5]
Output: 2
Explanation: All bulbs turned on, are blue at the moment 3, and 4 (index-0).

  

 1 class Solution:
 2     def numTimesAllBlue(self, light: List[int]) -> int:
 3         count = 0
 4         max1 = -1
 5         for i, key in enumerate(light):
 6             if key > max1:
 7                 max1 = key
 8             if max1 == i + 1:
 9                 count += 1
10         return count
11             

 

全部评论

相关推荐

牛客389580366号:读书的意义在于提升自己和他人吧,“阶级意识”是读书过程中的产出,“跨越阶级”是通过读书获得的能力
点赞 评论 收藏
分享
评论
点赞
收藏
分享
牛客网
牛客企业服务