python 
class Solution:
    def numberOfAlternatingGroups(self, colors: List[int]) -> int:
        count = 0
        # check the end colors 
        if len(colors) > 2:
            if colors[0] != colors[1] and colors[1] == colors[-1]:
                count += 1
            if colors[-2] != colors[-1] and colors[-2] == colors[0]:
                count += 1
        
        l, r = 0, 0
        k = 3
        for r in range(len(colors)):
            if r - l + 1 == 3:
                l += 1
            if r < len(colors) - 1 and colors[l] == colors[r+1] and colors[l] != colors[r]:
                count += 1

        return count

class Solution:
    def numberOfAlternatingGroups(self, colors: List[int]) -> int:
        n = len(colors) # keep the orginal length for latter
        count = 0

        # extend the colors array to simulate a circular pattern 
        colors.extend(colors[:2])

        for r in range(n): 
            if colors[r] == colors[r+2] and colors[r] != colors[r+1]:
                count += 1
                
        return count

python 
class Solution:
    def numberOfAlternatingGroups(self, colors: List[int], k: int) -> int:
        count = 0

        colors.extend(colors[:k-1]) # if k = 3 we add 2 numbers like the above question
        print(colors)

        L = 0
        for R in range(1, len(colors)):
            
            if colors[R] == colors[R-1]:
                L = R
            if R - L + 1 == k:
                count += 1
                L += 1

        return count

class Solution:
    def minOperations(self, nums: List[int]) -> int:
        n = len(nums)
        count = L =0
        for R in range(n - 2): 
            if nums[R] == 0:  
                nums[R] = 1 - nums[R]
                nums[R + 1] = 1 - nums[R + 1]
                nums[R + 2] = 1 - nums[R + 2]
                count += 1

            if nums[L] == 1: L+= 1
        
        for i in range(len(nums)):
            if nums[i] == 0:
                return -1
        return count