Input: nums = [10,5,2,6], k = 100 Output: 8 Explanation: The 8 subarrays that have product less than 100 are: , , , , [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.
Input: nums = [1,2,3], k = 0 Output: 0
1 <= nums.length <= 3 * 104 1 <= nums[i] <= 1000 0 <= k <= 106
The simplest way to think of this problem is taking cumulative product from an index up to the end of the array. Then check if any of those
products < k or not. We will have to do this for each index. Then, for any subarray
product < k, we can count the number of subarrays and find the result by adding up those counts.
But this approach will cost us O(N2) [O N square] runtime which is not efficient and we don't want that! So, let's not waste time on it. The better approach to solve this problem is to .... read more
Editorial source: foolishhungry blog