Description
The score of an array is defined as the product of its sum and its length.
For example, the score of [1, 2, 3, 4, 5] is (1 + 2 + 3 + 4 + 5) * 5 = 75.
Given a positive integer array nums and an integer k, return the number of non–empty subarrays of nums whose score is strictly less than k.
A subarray is a contiguous sequence of elements within an array.
Example 1:
Input: nums = [2,1,4,3,5], k = 10
Output: 6
Explanation:
The 6 subarrays having scores less than 10 are:
- [2] with score 2 * 1 = 2.
- [1] with score 1 * 1 = 1.
- [4] with score 4 * 1 = 4.
- [3] with score 3 * 1 = 3.
- [5] with score 5 * 1 = 5.
- [2,1] with score (2 + 1) * 2 = 6.
Note that subarrays such as [1,4] and [4,3,5] are not considered because their scores are 10 and 36 respectively, while we need scores strictly less than 10.
Example 2:
Input: nums = [1,1,1], k = 5
Output: 5
Explanation:
Every subarray except [1,1,1] has a score less than 5.
[1,1,1] has a score (1 + 1 + 1) * 3 = 9, which is greater than 5.
Thus, there are 5 subarrays having scores less than 5.
1 <= nums.length <= 10^5
1 <= nums[i] <= 10^5
1 <= k <= 10^15
Solution
Sliding window, when the score of the subarray is not less than k
, shrink the window. Since we need to calculate the number of the subarray, every time we expand the window, calculate the number of subarray with nums[j]
as the ending element.
o
(
n
)
o(n)
o(n)
Space complexity:
o
(
1
)
o(1)
o(1)
Code
class Solution:
def countSubarrays(self, nums: List[int], k: int) -> int:
cur_sum, res = 0, 0
i = 0
for j in range(len(nums)):
cur_sum += nums[j]
while cur_sum * (j - i + 1) >= k:
cur_sum -= nums[i]
i += 1
# j - i + 1 is the number of subarray with nums[j] as the ending element
res += j - i + 1
return res
原文地址:https://blog.csdn.net/sinat_41679123/article/details/134623612
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