<p>Given an array of integers <code>nums</code> and an integer <code>k</code>, return <em>the number of contiguous subarrays where the product of all the elements in the subarray is strictly less than </em><code>k</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [10,5,2,6], k = 100
<strong>Output:</strong> 8
<strong>Explanation:</strong> 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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3], k = 0
<strong>Output:</strong> 0
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 3 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 1000</code></li>
	<li><code>0 &lt;= k &lt;= 10<sup>6</sup></code></li>
</ul>

class Solution:
  def numSubarrayProductLessThanK(self, nums: list[int], k: int) -> int:
    if k <= 1:
      return 0

    ans = 0
    prod = 1

    j = 0
    for i, num in enumerate(nums):
      prod *= num
      while prod >= k:
        prod /= nums[j]
        j += 1
      ans += i - j + 1

    return ans
17900