Introduction

In this blog post, we will explore how to solve the LeetCode problem "713" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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: 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. Example 2: Input: nums = [1,2,3], k = 0 Output: 0 Constraints: 1 <= nums.length <= 3 * 104 1 <= nums[i] <= 1000 0 <= k <= 106

Explanation

Here's the solution to the problem:

• Sliding Window: Use a sliding window approach. Maintain a window of elements and expand it to the right.
• Product Calculation: Keep track of the product of the elements within the window. If the product becomes greater than or equal to k, shrink the window from the left until the product is less than k.

• Count Subarrays: For each valid window, the number of contiguous subarrays ending at the rightmost element is equal to the window size. Add this to the total count.

• Runtime Complexity: O(n) & Storage Complexity: O(1)

Code

    def numSubarrayProductLessThanK(nums: list[int], k: int) -> int:
    """
    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:
    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.

    Example 2:
    Input: nums = [1,2,3], k = 0
    Output: 0

    Constraints:
    1 <= nums.length <= 3 * 104
    1 <= nums[i] <= 1000
    0 <= k <= 106
    """
    if k <= 1:
        return 0

    left = 0
    product = 1
    count = 0

    for right in range(len(nums)):
        product *= nums[right]

        while product >= k:
            product /= nums[left]
            left += 1

        count += right - left + 1

    return count