Problem Description

You have a circle of tiles where each tile is colored either red or blue. The tiles are arranged in a circular formation, meaning the first and last tiles are neighbors.

Given:

• An array colors where each element represents a tile's color:
  • colors[i] == 0 means tile i is red
  • colors[i] == 1 means tile i is blue
• An integer k representing the size of groups to check

An alternating group is defined as k consecutive tiles in the circle where the colors alternate. This means every tile in the group (except the first and last) has a different color from both its left and right neighbors.

Your task is to count how many such alternating groups of size k exist in the circle.

For example, if you have tiles with colors [0, 1, 0, 1] and k = 3, you need to check all possible groups of 3 consecutive tiles:

• Group starting at index 0: [0, 1, 0] - alternates (0→1→0)
• Group starting at index 1: [1, 0, 1] - alternates (1→0→1)
• Group starting at index 2: [0, 1, 0] - alternates (0→1→0)
• Group starting at index 3: [1, 0, 1] - alternates (1→0→1)

Since the tiles form a circle, when checking the group starting at the last positions, you wrap around to the beginning of the array.

The solution uses a clever approach of "unfolding" the circular array by concatenating it with itself (creating a virtual array of length 2n), then using a sliding window technique to count valid alternating groups while avoiding duplicate counts.

Intuition

The key challenge with this problem is handling the circular nature of the tiles. When we check for alternating groups, groups can wrap around from the end of the array back to the beginning. This makes it tricky to check all possible groups efficiently.

One way to handle circular arrays is to "unfold" them. Imagine cutting the circle at one point and laying it out flat, then appending a copy of the same sequence right after it. This gives us a linear array of length 2n where any group of k consecutive elements in the original circle can be found as a contiguous subarray in this unfolded version.

For example, if our circle is [0, 1, 0, 1], unfolding it gives us [0, 1, 0, 1, 0, 1, 0, 1]. Now a group that wraps around (like the last tile and first two tiles) can be found without special wraparound logic.

To count alternating groups efficiently, we can track the current "streak" of alternating colors as we traverse the unfolded array. We maintain a counter cnt that:

• Increases by 1 when we see a color different from the previous one (alternating continues)
• Resets to 1 when we see the same color as the previous one (alternation breaks)

Whenever our streak length cnt is at least k, we've found a valid alternating group. But we need to be careful not to count the same group multiple times. Since we're traversing an unfolded array of length 2n, we only start counting groups after we've passed position n (the second half of our unfolded array). This ensures each group in the original circle is counted exactly once.

The beauty of this approach is that we only need a single pass through the unfolded array, checking each element once, giving us an efficient O(n) solution.

Learn more about Sliding Window patterns.

Solution Approach

The implementation uses the "unfolding" technique to handle the circular array efficiently. Here's how the solution works step by step:

1. Initialize Variables:

• n = len(colors): Store the length of the original array
• ans = 0: Counter for the number of alternating groups
• cnt = 0: Counter for the current streak of alternating colors

2. Traverse the Unfolded Array: We iterate through 2n positions using for i in range(n << 1) (where n << 1 is equivalent to n * 2). This simulates traversing the circular array twice.

3. Check for Alternation: At each position i:

• Calculate the actual index in the original array using modulo: i % n
• Compare the current color colors[i % n] with the previous color colors[(i - 1) % n]
• If i > 0 and the colors are the same, reset cnt = 1 (alternation breaks)
• Otherwise, increment cnt += 1 (alternation continues)

4. Count Valid Groups: The key line is ans += i >= n and cnt >= k:

• We only start counting when i >= n to avoid counting the same group twice
• We have a valid alternating group when cnt >= k
• The expression evaluates to 1 (True) when both conditions are met, 0 otherwise

5. Return the Result: After traversing all 2n positions, return ans which contains the total count of alternating groups.

Example Walkthrough: For colors = [0, 1, 0, 1] and k = 3:

• The unfolded sequence would be conceptually [0, 1, 0, 1, 0, 1, 0, 1]
• As we traverse, we track alternating streaks
• Starting from position n = 4, whenever we have a streak of at least 3, we increment our answer
• This correctly identifies all 4 alternating groups of size 3 in the circle

The time complexity is O(n) as we make a single pass through 2n elements, and the space complexity is O(1) as we only use a few variables regardless of input size.

Example Walkthrough

Let's walk through the solution with colors = [0, 1, 0, 0, 1] and k = 3.

Step 1: Conceptual Unfolding The circular array conceptually becomes: [0, 1, 0, 0, 1, 0, 1, 0, 0, 1]

Step 2: Traverse and Track Alternation We'll track cnt (alternation streak) and ans (valid groups found):

Step 3: Understanding the Count

• At i=5,6,7: We found alternating groups because cnt ≥ 3 and i ≥ 5
• These correspond to:
  • Group starting at index 3: [0, 1, 0] (indices 3→4→0 in circle)
  • Group starting at index 4: [1, 0, 1] (indices 4→0→1 in circle)
  • Group starting at index 0: [0, 1, 0] (indices 0→1→2)

Result: The algorithm correctly finds 3 alternating groups of size 3.

The key insight is that by traversing the "unfolded" array and only counting after position n, each valid group in the original circle is counted exactly once.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the array colors. The algorithm iterates through 2n elements (from 0 to 2n-1) in the for loop, performing constant-time operations in each iteration. The modulo operations i % n and (i - 1) % n, the comparison colors[i % n] == colors[(i - 1) % n], and the increment operations all take O(1) time. Therefore, the total time complexity is O(2n) = O(n).

The space complexity is O(1). The algorithm only uses a fixed amount of extra space for the variables n, ans, cnt, and i, regardless of the input size. No additional data structures that scale with the input are created.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Forgetting to Handle the Circular Nature Properly

One of the most common mistakes is treating the array as linear instead of circular. Developers might write code that only checks groups within the array bounds without considering wrap-around cases.

Incorrect approach:

# This misses groups that wrap around
for i in range(n - k + 1):
    # Check if group starting at i is alternating
    is_alternating = True
    for j in range(i + 1, i + k):
        if colors[j] == colors[j - 1]:
            is_alternating = False
            break

Solution: Use the array unfolding technique (iterating through 2n positions) or explicitly handle wrap-around with modulo operations.

2. Double Counting Groups

When unfolding the array, there's a risk of counting the same group multiple times if you start counting too early.

Incorrect approach:

for i in range(n * 2):
    # Starting to count from i = 0 causes duplicates
    if current_alternating_length >= k:
        result += 1

Solution: Only start counting when i >= n. This ensures each group is counted exactly once:

if i >= n and current_alternating_length >= k:
    result += 1

3. Incorrectly Resetting the Alternating Counter

A subtle bug occurs when developers reset the counter to 0 instead of 1 when the alternation breaks.

Incorrect approach:

if i > 0 and colors[current_idx] == colors[previous_idx]:
    current_alternating_length = 0  # Wrong! Should be 1

Solution: When alternation breaks, the current tile still starts a new potential alternating sequence, so set the counter to 1:

if i > 0 and colors[current_idx] == colors[previous_idx]:
    current_alternating_length = 1  # Correct

4. Off-by-One Errors in Index Calculation

When working with circular arrays, index calculations can be tricky, especially for the previous element.

Incorrect approach:

previous_idx = (i - 1) % n  # Can give unexpected results when i = 0

Solution: Add proper boundary checks or ensure the logic handles all edge cases:

if i > 0:  # Only check previous element after the first iteration
    previous_idx = (i - 1) % n
    if colors[current_idx] == colors[previous_idx]:
        current_alternating_length = 1

5. Not Understanding the Problem Definition

Some might misinterpret "alternating group" as requiring the group to start and end with different colors, rather than having each interior element differ from its neighbors.

Solution: Carefully verify that the alternation check compares consecutive elements, not just the pattern of the group. A valid alternating group of size k means there are k consecutive tiles where each adjacent pair has different colors.