The Question
Coding

Subarray Product Less Than K

Given an array of positive 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. Constraints: - 1 <= nums.length <= 3 * 10^4 - 1 <= nums[i] <= 1000 - 0 <= k <= 10^6 Example: Input: nums = [10, 5, 2, 6], k = 100 Output: 8 Explanation: The 8 subarrays are: [10], [5], [2], [6], [10, 5], [5, 2], [2, 6], [5, 2, 6]. Note that [10, 5, 2] is not included as its product is 100, which is not strictly less than k.
Java
Sliding Window
Two Pointers
Questions & Insights

Clarifying Questions

Are all elements in the input array strictly positive?
Assumption: Based on the constraints, nums[i] \ge 1. This is crucial because it ensures the product is monotonically increasing as the window expands.
What is the range of $k$?
Assumption: 0 \le k \le 10^6. If k \le 1, since all array elements are \ge 1, no subarray product can be strictly less than k.
Can the product of a window exceed the maximum value of a 32-bit integer?
Assumption: The maximum value of k is 10^6 and the maximum element is 10^3. In the sliding window, the product will be reduced as soon as it exceeds or equals k. The maximum temporary product would be roughly k \times \max(nums[i]), which is 10^9. This fits within a standard 32-bit signed int (up to 2.1 \times 10^9), but using a long for the product is a safer practice to prevent overflow in different variations of this problem.

Thinking Process

Monotonicity: Since all numbers are positive, adding an element to a subarray will always increase (or maintain) its product, and removing an element will always decrease (or maintain) its product. This property allows us to use a Sliding Window (Two Pointers) approach.
Window Management: Maintain a window [left, right] and a running product. For every right pointer position, we find the largest valid window ending at right by incrementing the left pointer until the product < k.
Counting Subarrays: For a fixed right index, if the window [left, right] is valid, then all subarrays starting from left, left+1, ..., right and ending at right are also valid. The number of such subarrays is exactly right - left + 1.
Edge Cases: If k \le 1, it is mathematically impossible for any product of integers \ge 1 to be strictly less than k. We should return 0 immediately.
Implementation Breakdown

Problem Set

Functional Requirements:
Given an array of integers nums and an integer k.
Return the count of contiguous subarrays where the product of all elements is strictly less than k.
Constraints:
1 \le nums.length \le 3 \cdot 10^4
1 \le nums[i] \le 1000
0 \le k \le 10^6
Time Complexity: O(n)
Space Complexity: O(1)

Approach

Algorithm: Sliding Window (Two Pointers)
Data Structure: Array
Complexity:
Time: O(n), where n is the length of the array. Each pointer travels across the array at most once.
Space: O(1), as we only store a few primitive variables.

Implementation

Wrap Up

Advanced Topics

Handling Zeros: If the array contained zeros, the sliding window logic would break (division by zero and non-monotonicity). One would need to split the array at zero positions or use a different counting logic.
Large Products / Precision: If k or elements were significantly larger (exceeding long capacity), we could use logarithms: \log(a \cdot b) = \log(a) + \log(b). The condition product < k becomes \sum \log(nums[i]) < \log(k). However, one must be careful with floating-point precision issues.
Parallelization: This specific problem is inherently sequential due to the contiguous subarray constraint. However, for extremely large datasets, one could use a Divide and Conquer approach (similar to counting inversions) or process chunks and handle subarrays crossing boundaries separately, though this increases complexity significantly.