Introduction

In this blog post, we will explore how to solve the LeetCode problem "2488" 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 array nums of size n consisting of distinct integers from 1 to n and a positive integer k. Return the number of non-empty subarrays in nums that have a median equal to k. Note: The median of an array is the middle element after sorting the array in ascending order. If the array is of even length, the median is the left middle element. For example, the median of [2,3,1,4] is 2, and the median of [8,4,3,5,1] is 4. A subarray is a contiguous part of an array. Example 1: Input: nums = [3,2,1,4,5], k = 4 Output: 3 Explanation: The subarrays that have a median equal to 4 are: [4], [4,5] and [1,4,5]. Example 2: Input: nums = [2,3,1], k = 3 Output: 1 Explanation: [3] is the only subarray that has a median equal to 3. Constraints: n == nums.length 1 <= n <= 105 1 <= nums[i], k <= n The integers in nums are distinct.

Explanation

Here's the breakdown of the approach, complexity, and Python code for solving the median subarray problem:

• Key Idea: Iterate through each element of the input array nums. For each element, expand outward to consider all possible subarrays containing that element. Check if the median of the current subarray is equal to k.
• Optimization: To efficiently calculate the median of a subarray, we can avoid explicitly sorting the subarray each time. Instead, we count the number of elements less than k and greater than k in the subarray. The median is k if and only if the number of elements less than k is equal to or one greater than the number of elements greater than k.

• Efficient Counting: Maintain counts of elements less than k and greater than k as you expand the subarray.

• Complexity: Runtime: O(n^2), Storage: O(1)

Code

    def count_subarrays(nums, k):
    """
    Counts the number of non-empty subarrays in nums that have a median equal to k.
    """
    n = len(nums)
    count = 0

    for i in range(n):
        for j in range(i, n):
            subarray = nums[i:j+1]
            subarray_len = len(subarray)

            less_than_k = 0
            greater_than_k = 0

            for num in subarray:
                if num < k:
                    less_than_k += 1
                elif num > k:
                    greater_than_k += 1

            if (less_than_k == greater_than_k and k in subarray and subarray_len % 2 != 0):
                 count +=1
            elif (less_than_k - greater_than_k == 1 and k in subarray and subarray_len % 2 == 0):
                count += 1
            elif (less_than_k == greater_than_k and k in subarray and subarray_len % 2 == 0):
                continue
            elif (subarray_len == 1 and k in subarray):
                count +=1


    return count