Description
Given an array of integers nums and an integer k, return the number of contiguous subarrays where the product of all the elements in the subarray is strictly less than k.
Example 1:
Input: nums = [10,5,2,6], k = 100
Output: 8
Explanation: The 8 subarrays that have product less than 100 are:
[10], [5], [2], [6], [10, 5], [5, 2], [2, 6], [5, 2, 6]
Note that [10, 5, 2] is not included as the product of 100 is not strictly less than k.
Example 2:
Input: nums = [1,2,3], k = 0
Output: 0
1 <= nums.length <= 3 * 10^4
1 <= nums[i] <= 1000
0 <= k <= 10^6
Solution
Similar to 2302. Count Subarrays With Score Less Than K, every time when expanding the window, we introduce j - i + 1 new subarrays (with the nums[j] as the ending element)
o
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n
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o(n)
o(n)
Space complexity:
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o(1)
o(1)
Code
class Solution:
def numSubarrayProductLessThanK(self, nums: List[int], k: int) -> int:
cur_prod = 1
i = 0
res = 0
for j in range(len(nums)):
cur_prod *= nums[j]
while i <= j and cur_prod >= k:
cur_prod /= nums[i]
i += 1
res += j - i + 1
return res
原文地址:https://blog.csdn.net/sinat_41679123/article/details/134623616
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