Naive Solution

The brute force approach is to iterate through the colors array and check for each group of three contiguous tiles if they form an alternating group. Since the tiles are arranged in a circle, we need to handle the wrap-around case where the first and last tiles are considered adjacent.

Algorithm

1. Iterate through the colors array from index 0 to n-1, where n is the length of the array.
2. For each index i, check if the tiles at indices i, (i+1)%n, and (i+2)%n form an alternating group.
3. A group forms an alternating group if colors[i] != colors[(i+1)%n] and colors[(i+1)%n] != colors[(i+2)%n].
4. Count the number of alternating groups.

Code

def count_alternating_groups_naive(colors):
    n = len(colors)
    count = 0
    for i in range(n):
        if colors[i] != colors[(i + 1) % n] and colors[(i + 1) % n] != colors[(i + 2) % n]:
            count += 1
    return count

Time and Space Complexity

• Time Complexity: O(n), where n is the number of tiles, as we iterate through the array once.
• Space Complexity: O(1), as we use a constant amount of extra space.

Optimal Solution

The optimal solution is essentially the same as the naive solution because we must inspect each possible group of three contiguous tiles to determine if it's an alternating group. Therefore, there's no way to significantly improve upon the naive solution in terms of time complexity.

Algorithm

(Same as the naive solution)

1. Iterate through the colors array from index 0 to n-1, where n is the length of the array.
2. For each index i, check if the tiles at indices i, (i+1)%n, and (i+2)%n form an alternating group.
3. A group forms an alternating group if colors[i] != colors[(i+1)%n] and colors[(i+1)%n] != colors[(i+2)%n].
4. Count the number of alternating groups.

Code

def count_alternating_groups_optimal(colors):
    n = len(colors)
    count = 0
    for i in range(n):
        if colors[i] != colors[(i + 1) % n] and colors[(i + 1) % n] != colors[(i + 2) % n]:
            count += 1
    return count

Time and Space Complexity

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

Edge Cases

• Empty Array: The problem statement says the length is always at least 3, so no check is needed.
• All tiles are the same color: The function should return 0.
• Alternating Colors: The function should return n.

Summary

Both the naive and optimal solutions have a time complexity of O(n) and a space complexity of O(1). The key is to correctly handle the circular nature of the array by using the modulo operator (%).