Given an array nums
sorted in non-decreasing order, return the maximum between the number of positive integers and the number of negative integers.
- In other words, if the number of positive integers in
nums
ispos
and the number of negative integers isneg
, then return the maximum ofpos
andneg
.
Note that 0
is neither positive nor negative.
Input: nums = [-2,-1,-1,1,2,3] Output: 3 Explanation: There are 3 positive integers and 3 negative integers. The maximum count among them is 3.
Input: nums = [-3,-2,-1,0,0,1,2] Output: 3 Explanation: There are 2 positive integers and 3 negative integers. The maximum count among them is 3.
Input: nums = [5,20,66,1314] Output: 4 Explanation: There are 4 positive integers and 0 negative integers. The maximum count among them is 4.
1 <= nums.length <= 2000
-2000 <= nums[i] <= 2000
nums
is sorted in a non-decreasing order.
Follow up: Can you solve the problem in O(log(n))
time complexity?
class Solution:
def maximumCount(self, nums: List[int]) -> int:
pos = bisect.bisect_left(nums, 0)
neg = len(nums) - bisect.bisect_right(nums, 0)
return max(pos, neg)