leetcode 713. Subarray Product Less Than K

Prefix product or two pointer can both solve this problem.
But don't forget to take a look at the note

Note:

0 < nums.length <= 50000.
0 < nums[i] < 1000.
0 <= k < 10^6.

which means the product can be as large as 10^150000. So two pointer is a better choice because you don't have to deal with big integer problem.
For each number, add up the count of subarray who's product < k and ends with the number.

class Solution {
    public int numSubarrayProductLessThanK(int[] nums, int k) {
        if (k <= 1) return 0;
        int product = 1;
        int left = 0, right = 0;
        int N = nums.length;
        int ret = 0;
        while (right < N) {
            product *= nums[right++];
            while (left < right && product >= k) {
                product /= nums[left++];
            }
            ret += (right - left);
        }
        return ret;
    }
}