# [3208\. Alternating Groups II](https://leetcode.com/problems/alternating-groups-ii/)

1. **Extend the Array**: To handle the circular nature of the array, extend the `colors` array by appending the first `k-1` elements to its end. This allows us to easily check for alternating groups that wrap around the end of the array.

2. **Initialize Counters**:

- `res` to store the number of alternating groups.
- `cnt` to count the length of the current alternating group.
3. **Iterate Through the Extended Array**:

- For each tile, check if its color is different from the previous tile.
- If it is, increment the `cnt` counter.
- If it isn't, reset the `cnt` counter to 1.
- If the `cnt` counter reaches `k`, increment the `res` counter as it indicates the end of an alternating group.
4. **Return the Result**: Finally, return the `res` counter, which represents the number of alternating groups of length `k`.

:::spoiler Solution

class Solution {
public:
    int numberOfAlternatingGroups(vector<int>& colors, int k)
    {
        for (int i = 0; i < k - 1; ++i)
        {
            colors.push_back(colors[i]);
        }

        int res = 0;

        // cnt 用來記錄長度是否達到 k
        int cnt = 1;

        for (int i = 1; i < colors.size(); ++i)
        {
            // 如果是交錯的格子，則 cnt++
            if (colors[i] != colors[i - 1])
            {
                ++cnt;
            }
            else
            {
                // 如果不是交錯的格子，則 reset cnt= 1
                cnt = 1;
            }
            // 如果 cnt 達到 k，則結果 + 1
            if (cnt >= k) ++res;
        }
        return res;        
    }
};

- 時間複雜度：$O(n + k - 1)$
- 空間複雜度：$O(n + k)$
:::