# Solving Leetcode Interviews in Seconds with AI: Alternating Groups I


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "3206" 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
	> There is a circle of red and blue tiles. You are given an array of integers colors. The color of tile i is represented by colors[i]:  colors[i] == 0 means that tile i is red. colors[i] == 1 means that tile i is blue.  Every 3 contiguous tiles in the circle with alternating colors (the middle tile has a different color from its left and right tiles) is called an alternating group. Return the number of alternating groups. Note that since colors represents a circle, the first and the last tiles are considered to be next to each other.   Example 1:  Input: colors = [1,1,1] Output: 0 Explanation:   Example 2:  Input: colors = [0,1,0,0,1] Output: 3 Explanation:  Alternating groups:     Constraints:  3 <= colors.length <= 100 0 <= colors[i] <= 1  

	# Explanation
	Here's the breakdown:

*   **Circular Handling:** Treat the array as circular by considering the first and last elements as neighbors.
*   **Alternating Group Detection:** Iterate through the array, checking each group of three adjacent tiles for the alternating color pattern.
*   **Counting:** Increment a counter each time an alternating group is found.

*   **Time Complexity:** O(n), where n is the number of tiles. **Space Complexity:** O(1).

	
	# Code
	```python
	def alternating_tiles(colors):
    """
    Counts the number of alternating groups in a circular arrangement of colored tiles.

    Args:
        colors: A list of integers representing the colors of the tiles (0 for red, 1 for blue).

    Returns:
        The number of alternating groups.
    """

    n = len(colors)
    count = 0

    for i in range(n):
        # Get the colors of the three tiles
        prev_index = (i - 1 + n) % n  # Handle circularity for previous tile
        next_index = (i + 1) % n      # Handle circularity for next tile

        prev_color = colors[prev_index]
        current_color = colors[i]
        next_color = colors[next_index]

        # Check for alternating pattern
        if prev_color != current_color and next_color != current_color:
            count += 1

    return count
	```