Introduction

In this blog post, we will explore how to solve the LeetCode problem "2962" 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

You are given an integer array nums and a positive integer k. Return the number of subarrays where the maximum element of nums appears at least k times in that subarray. A subarray is a contiguous sequence of elements within an array. Example 1: Input: nums = [1,3,2,3,3], k = 2 Output: 6 Explanation: The subarrays that contain the element 3 at least 2 times are: [1,3,2,3], [1,3,2,3,3], [3,2,3], [3,2,3,3], [2,3,3] and [3,3]. Example 2: Input: nums = [1,4,2,1], k = 3 Output: 0 Explanation: No subarray contains the element 4 at least 3 times. Constraints: 1 <= nums.length <= 105 1 <= nums[i] <= 106 1 <= k <= 105

Explanation

Here's the breakdown of the solution:

• Identify Maximum Element: Find the maximum element in the input array nums. This will be the target element for counting occurrences within subarrays.
• Sliding Window: Use a sliding window approach to efficiently iterate through all possible subarrays. The window expands to the right, and we maintain a count of the maximum element within the current window.

• Count Valid Subarrays: If the count of the maximum element within the window reaches at least k, we increment the result by the number of subarrays ending at the current position.

• Runtime Complexity: O(n), where n is the length of the nums array.

• Storage Complexity: O(1)

Code

    def count_subarrays(nums, k):
    """
    Counts the number of subarrays where the maximum element appears at least k times.

    Args:
        nums: An integer array.
        k: A positive integer.

    Returns:
        The number of subarrays where the maximum element of nums appears at least k times.
    """

    max_num = max(nums)
    n = len(nums)
    count = 0

    for i in range(n):
        max_count = 0
        for j in range(i, n):
            if nums[j] == max_num:
                max_count += 1
            if max_count >= k:
                count += (n - j)
                break  # Optimization: No need to continue if already met k

    return count