# Solving Leetcode Interviews in Seconds with AI: Three Consecutive Odds


	# 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](https://www.chatmagic.app), 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
	```python
	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
	```