Introduction

In this blog post, we will explore how to solve the LeetCode problem "2148" 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 integer array nums, return the number of elements that have both a strictly smaller and a strictly greater element appear in nums. Example 1: Input: nums = [11,7,2,15] Output: 2 Explanation: The element 7 has the element 2 strictly smaller than it and the element 11 strictly greater than it. Element 11 has element 7 strictly smaller than it and element 15 strictly greater than it. In total there are 2 elements having both a strictly smaller and a strictly greater element appear in nums. Example 2: Input: nums = [-3,3,3,90] Output: 2 Explanation: The element 3 has the element -3 strictly smaller than it and the element 90 strictly greater than it. Since there are two elements with the value 3, in total there are 2 elements having both a strictly smaller and a strictly greater element appear in nums. Constraints: 1 <= nums.length <= 100 -105 <= nums[i] <= 105

Explanation

Here's a breakdown of the solution:

• Identify Extremes: Find the minimum and maximum values in the array.

• Count Valid Elements: Iterate through the array and count elements that are strictly greater than the minimum and strictly smaller than the maximum.

• Runtime Complexity: O(n), where n is the length of the array. Storage Complexity: O(1).

Code

    def count_elements(nums):
    """
    Counts the number of elements in an array that have both a strictly smaller and a strictly greater element.

    Args:
        nums: An integer array.

    Returns:
        The number of elements that have both a strictly smaller and a strictly greater element.
    """

    if not nums or len(nums) < 3:
        return 0

    min_val = nums[0]
    max_val = nums[0]

    for num in nums:
        min_val = min(min_val, num)
        max_val = max(max_val, num)

    count = 0
    for num in nums:
        if num > min_val and num < max_val:
            count += 1

    return count