Introduction

In this blog post, we will explore how to solve the LeetCode problem "3151" 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

An array is considered special if the parity of every pair of adjacent elements is different. In other words, one element in each pair must be even, and the other must be odd. You are given an array of integers nums. Return true if nums is a special array, otherwise, return false. Example 1: Input: nums = [1] Output: true Explanation: There is only one element. So the answer is true. Example 2: Input: nums = [2,1,4] Output: true Explanation: There is only two pairs: (2,1) and (1,4), and both of them contain numbers with different parity. So the answer is true. Example 3: Input: nums = [4,3,1,6] Output: false Explanation: nums[1] and nums[2] are both odd. So the answer is false. Constraints: 1 <= nums.length <= 100 1 <= nums[i] <= 100

Explanation

Here's the breakdown of the solution:

• Iterate and Check Parity: The core idea is to iterate through the input array nums, examining each adjacent pair of elements.
• Parity Difference: For each pair, determine if their parities are different. If any pair has the same parity, the array is not special.

• Handle Single Element: The edge case of a single-element array is handled by returning True.

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

• Storage Complexity: O(1) - constant extra space.

Code

    def is_special_array(nums):
    """
    Checks if an array is special, meaning the parity of every pair of adjacent elements is different.

    Args:
        nums: A list of integers.

    Returns:
        True if the array is special, False otherwise.
    """
    n = len(nums)
    if n <= 1:
        return True

    for i in range(n - 1):
        if nums[i] % 2 == nums[i + 1] % 2:
            return False

    return True