# 713 Subarray Product Less Than K

## 713. [Subarray Product Less Than K](https://leetcode.com/problems/subarray-product-less-than-k/description/)

## 1. Question

Your are given an array of positive integers`nums`.

Count and print the number of (contiguous) subarrays where the product of all the elements in the subarray is less than`k`.

**Example 1:**

```
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.
```

**Note:**

`0 < nums.length <= 50000`.

`0 < nums[i] < 1000`.

`0 <= k < 10^6`.

## 2. Implementation

**(1) Two Pointers**

思路: 利用滑动窗口的思想，维护满足条件的两个pointer start和end。注意的是每当我们在右边增加一个数时，我们引入了end - start + 1个新的subarray。比如假设我们原来有\[1, 2]， 当我们再右边增加3时，我们比原来多了3个subarray，分别是:

\[3], \[1,3], \[2,3]

```java
class Solution {
    public int numSubarrayProductLessThanK(int[] nums, int k) {
        if (k <= 1) {
            return 0;
        }

        int prod = 1;
        int res = 0;
        for (int start = 0, end = 0; end < nums.length; end++) {
            prod *= nums[end];

            while (prod >= k) {
                prod /= nums[start];
                ++start;
            }
            res += end - start + 1;
        }
        return res;
    }
}
```

## 3. Time & Space Complexity

Two Pointers: 时间复杂度O(n), 空间复杂度O(1)