# 995. Minimum Number of K Consecutive Bit Flips
"""
given a binary array 'nums', and an integer 'k'. a k-bit flip is choosing a
subarray of length 'k' from 'nums' 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. a subarray is a contiguous part of an array.
"""
class Solution(object):
def minKBitFlips(self, nums, k):
"""
:type nums: List[int]
:type k: int
:rtype: int
"""
curr, ans, n = 0, 0, len(nums)
for i in range(len(nums)):
if i >= k and nums[i - k] > 1:
nums[i - k] -= 2
curr -= 1
if curr & 1 ^ nums[i] == 0:
if i + k > len(nums):
return -1
nums[i] += 2
curr += 1
ans += 1
return ans
if __name__ == "__main__":
obj = Solution()
print(obj.minKBitFlips(nums=[0, 1, 0], k=1))
print(obj.minKBitFlips(nums=[1, 1, 0], k=2))
print(obj.minKBitFlips(nums=[0, 0, 0, 1, 0, 1, 1, 0], k=3))
