If the product of elements from l to r is < k, then all subarrays ending at r starting between l and r also satisfy the condition. The number of such subarrays is r - l + 1.

1. If k <= 1, return 0.
2. l = 0, prod = 1, res = 0.
3. For each r, prod *= nums[r].
4. While prod >= k (and l <= r):
   • prod /= nums[l].
   • l += 1.
5. res += (r - l + 1).
6. Return res.

• Time Complexity: O(N).
• Space Complexity: O(1).

def num_subarray_product_less_than_k(nums, k):
    if k <= 1: return 0
    res = 0
    prod = 1
    l = 0
    for r in range(len(nums)):
        prod *= nums[r]
        while prod >= k:
            prod /= nums[l]
            l += 1
        res += (r - l + 1)
    return res