In an array A containing only 0s and 1s, a K-bit flip consists of choosing a (contiguous) subarray of length K and simultaneously changing every 0 in the subarray to 1, and every 1 in the subarray to 0.
Return the minimum number of K-bit flips required so that there is no 0 in the array.  If it is not possible, return -1.
Example 1:
Input: A = [0,1,0], K = 1
Output: 2
Explanation: Flip A[0], then flip A[2].
Example 2:
Input: A = [1,1,0], K = 2
Output: -1
Explanation: No matter how we flip subarrays of size 2, we can't make the array become [1,1,1].
Example 3:
Input: A = [0,0,0,1,0,1,1,0], K = 3
Output: 3
Explanation:
Flip A[0],A[1],A[2]: A becomes [1,1,1,1,0,1,1,0]
Flip A[4],A[5],A[6]: A becomes [1,1,1,1,1,0,0,0]
Flip A[5],A[6],A[7]: A becomes [1,1,1,1,1,1,1,1]
Note:
1 <= A.length <= 30000
1 <= K <= A.length
class Solution:
    def minKBitFlips(self, A: List[int], K: int) -> int:
        flip_time=0
        n=len(A)
        close=[False for i in range(n)]
        #print(close)
        res=0
        for i in range(n):
            if close[i]:
                flip_time-=1
            if (A[i]==0 and flip_time%2==0) or (A[i]==1 and flip_time%2==1):
                flip_time+=1
                res+=1
                if i+K>n: return -1
                if i+K<n:
                    close[i+K]=True
        return res
                
                
        

class Solution:
    def minKBitFlips(self, A: List[int], K: int) -> int:
        flip_time=0
        n=len(A)
        close=[False for i in range(n)]
        #print(close)
        res=0
        for i in range(n):
            if close[i]:
                flip_time-=1
            if (A[i]==0 and flip_time%2==0) or (A[i]==1 and flip_time%2==1):
                flip_time+=1
                res+=1
                if i+K>n: return -1
                if i+K<n:
                    close[i+K]=True
        return res
                
                
        
