Subarray Product Less Than K
<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>
Array; Binary Search; Prefix Sum; Sliding Window
class Solution {
public:
    int numSubarrayProductLessThanK(vector<int>& nums, int k) {
        int ans = 0, l = 0;
        int p = 1;
        for (int r = 0; r < nums.size(); ++r) {
            p *= nums[r];
            while (l <= r && p >= k) {
                p /= nums[l++];
            }
            ans += r - l + 1;
        }
        return ans;
    }
};
