Introduction

In this blog post, we will explore how to solve the LeetCode problem "1550" 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 arr, return true if there are three consecutive odd numbers in the array. Otherwise, return false. Example 1: Input: arr = [2,6,4,1] Output: false Explanation: There are no three consecutive odds. Example 2: Input: arr = [1,2,34,3,4,5,7,23,12] Output: true Explanation: [5,7,23] are three consecutive odds. Constraints: 1 <= arr.length <= 1000 1 <= arr[i] <= 1000

Explanation

Here's an efficient solution to determine if an array contains three consecutive odd numbers:

• Iterate and Check: Iterate through the array, checking if the current element and the next two elements (if they exist within the array bounds) are all odd.
• Early Exit: If three consecutive odd numbers are found, immediately return True.

• Default Return: If the loop completes without finding such a sequence, return False.

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

Code

    def three_consecutive_odds(arr: list[int]) -> bool:
    """
    Given an integer array arr, return true if there are three consecutive odd numbers in the array.
    Otherwise, return false.

    Example 1:
    Input: arr = [2,6,4,1]
    Output: false
    Explanation: There are no three consecutive odds.

    Example 2:
    Input: arr = [1,2,34,3,4,5,7,23,12]
    Output: true
    Explanation: [5,7,23] are three consecutive odds.

    Constraints:
    1 <= arr.length <= 1000
    1 <= arr[i] <= 1000
    """
    for i in range(len(arr) - 2):
        if arr[i] % 2 != 0 and arr[i + 1] % 2 != 0 and arr[i + 2] % 2 != 0:
            return True
    return False