Leetcode 713 Subarray Product Less Than K

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.

Input: nums = [1,2,3], k = 0
Output: 0

• Input: nums = [10, 5, 2, 6], k = 100

======== (End for loop, r = 0 snapshot) ============= [10, 5, 2, 6] r l prod = 10 cnt += 1

======== (End for loop, r = 1 snapshot) ============= [10, 5, 2, 6] r l prod = 50 cnt += 2

======== (r = 2, prod >= k snapshot) ================ [10, 5, 2, 6] r l
prod = 100, >= k, keep moving l till < k

======== (End for loop, r = 2 snapshot) ============= [10, 5, 2, 6] r l
prod = 10 cnt += 2

======== (End for loop, r = 3 snapshot) ============= [10, 5, 2, 6] r l
prod = 60 cnt += 3

class Solution(object):
    def numSubarrayProductLessThanK(self, nums: List[int], k: int) -> int:        
        left, prod, count = 0, 1, 0
            
        for right in range(len(nums)):
            prod *= nums[right]            
        
            while prod >= k and left <= right:                    
                prod /= nums[left]
                left += 1                        
            count += right - left + 1                
        return count