# 713. Subarray Product Less Than K

**713。Subarray Product Less Than K**

Given an array of integers `nums` and an integer `k`, return *the number of contiguous subarrays where the product of all the elements in the subarray is strictly less than* `k`.

**Example 1:**

<pre><code><strong>Input: nums = [10,5,2,6], k = 100
</strong><strong>Output: 8
</strong><strong>Explanation: The 8 subarrays that have product less than 100 are:
</strong>[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.
</code></pre>

**Example 2:**

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

**Constraints:**

* `1 <= nums.length <= 3 * 104`
* `1 <= nums[i] <= 1000`
* `0 <= k <= 106`

***My Solutions:***

```java
class Solution {
    public int numSubarrayProductLessThanK(int[] nums, int k) {
        if (k <= 1) return 0;
        int prod = 1, ans = 0, l = 0;
        for (int r = 0; r < nums.length; r++) {
            prod *= nums[r];
            // 如果结果超过k，把window的左边左移
            while (prod >= k) {
                prod  /= nums[l];
                l++;
            }
            // 这一步关键。可以组成的subarray数量是r - l + 1
            // e.g. l=0,r=0, ans+=1, [10]
            // e.g. l=0,r=1, ans+=2, [10,5],[5]
            // e.g. l=1,r=2, ans+=2, [5,2],[2]
            // e.g. l=1,r=3, ans+=3, [5,2,6],[2,6],[6] 
            ans += r - l + 1;
        }

        return ans;
    }
}
```