713. Subarray Product Less Than K

https://leetcode.com/problems/subarray-product-less-than-k/

Medium

Description

Your are given an array of positive integers nums.

Count and print the number of (contiguous) subarrays where the product of all the elements in the subarray is less than k.

Example

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.

Solution

class Solution:
    def numSubarrayProductLessThanK(self, nums: List[int], k: int) -> int:

      if k<= 1:
        return 0
        
      count = 0
      start = 0
      end = 0
      sum = 1

      while end < len(nums):

        v = nums[end]
        sum *= v

        while sum >= k:
          sum /= nums[start]
          start += 1

        count += end - start + 1
        end += 1
      
      return count

Tips

• Two Pointer
• Sliding Window

Note

以原題目[10,5,2,6]解釋
當[5,2,6]成立時,包含的[5,2,6],[2,6],[6]皆成立
end-start+1就是所有集合的數量