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+# 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))