Subarray product less than K [Sliding window]

Time: O(N); Space: O(1); medium

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 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.

Notes:

  • 0 < nums.length <= 50000.

  • 0 < nums[i] < 1000.

  • 0 <= k < 10^6.

[1]:

class Solution1(object):
    def numSubarrayProductLessThanK(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: int
        """
        if k <= 1: return 0
        result, start, prod = 0, 0, 1
        for i, num in enumerate(nums):
            prod *= num
            while prod >= k:
                prod /= nums[start]
                start += 1
            result += i-start+1
        return result

[2]:

s = Solution1()
nums = [10, 5, 2, 6]
k = 100
assert s.numSubarrayProductLessThanK(nums, k) == 8