Introduction

In this blog post, we will explore how to solve the LeetCode problem "2302" 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

The score of an array is defined as the product of its sum and its length. For example, the score of [1, 2, 3, 4, 5] is (1 + 2 + 3 + 4 + 5) 5 = 75. Given a positive integer array nums and an integer k, return the number of non-empty subarrays of nums whose score is strictly less than k. A subarray is a contiguous sequence of elements within an array. Example 1: Input: nums = [2,1,4,3,5], k = 10 Output: 6 Explanation: The 6 subarrays having scores less than 10 are: - [2] with score 2 1 = 2. - [1] with score 1 1 = 1. - [4] with score 4 1 = 4. - [3] with score 3 1 = 3. - [5] with score 5 1 = 5. - [2,1] with score (2 + 1) 2 = 6. Note that subarrays such as [1,4] and [4,3,5] are not considered because their scores are 10 and 36 respectively, while we need scores strictly less than 10. Example 2: Input: nums = [1,1,1], k = 5 Output: 5 Explanation: Every subarray except [1,1,1] has a score less than 5. [1,1,1] has a score (1 + 1 + 1) 3 = 9, which is greater than 5. Thus, there are 5 subarrays having scores less than 5. Constraints: 1 <= nums.length <= 105 1 <= nums[i] <= 105 1 <= k <= 1015

Explanation

Here's the solution to the problem:

• Sliding Window Approach: Use a sliding window to efficiently calculate the sum of subarrays. The window expands to the right, and contracts from the left when the score exceeds the limit k.

• Count Valid Subarrays: For each valid window, the number of subarrays ending at the current right endpoint is simply the window's length. Accumulate these counts to find the total.

• Handle Integer Overflow: Since k can be up to 10¹⁵, intermediate calculations (sum * length) can potentially overflow. Use 64-bit integers to prevent this.

• Runtime Complexity: O(n), where n is the length of the nums array because we iterate through the array once with the sliding window.

• Storage Complexity: O(1), as we use a constant amount of extra space.

Code

    def count_subarrays(nums, k):
    """
    Counts the number of non-empty subarrays of nums whose score is strictly less than k.

    Args:
        nums: A list of positive integers.
        k: An integer.

    Returns:
        The number of subarrays with score less than k.
    """

    count = 0
    left = 0
    current_sum = 0

    for right in range(len(nums)):
        current_sum += nums[right]

        while current_sum * (right - left + 1) >= k and left <= right:
            current_sum -= nums[left]
            left += 1

        count += (right - left + 1)

    return count