## Algorithm

Problem Name: 995. Minimum Number of K Consecutive Bit Flips

You are 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.

Example 1:

```Input: nums = [0,1,0], k = 1
Output: 2
Explanation: Flip nums[0], then flip nums[2].
```

Example 2:

```Input: nums = [1,1,0], k = 2
Output: -1
Explanation: No matter how we flip subarrays of size 2, we cannot make the array become [1,1,1].
```

Example 3:

```Input: nums = [0,0,0,1,0,1,1,0], k = 3
Output: 3
Explanation:
Flip nums[0],nums[1],nums[2]: nums becomes [1,1,1,1,0,1,1,0]
Flip nums[4],nums[5],nums[6]: nums becomes [1,1,1,1,1,0,0,0]
Flip nums[5],nums[6],nums[7]: nums becomes [1,1,1,1,1,1,1,1]
```

Constraints:

• `1 <= nums.length <= 105`
• `1 <= k <= nums.length`

## Code Examples

### #1 Code Example with Javascript Programming

```Code - Javascript Programming```

``````
const minKBitFlips = function(nums, k) {
let cur = 0, res = 0
const n = nums.length
for(let i = 0; i  <  n; i++) {
if(i >= k && nums[i - k] === 2) cur--
if(cur % 2 === nums[i]) {
if(i + k > n) return -1
nums[i] = 2
cur++
res++
}
}
return res
};
``````
Copy The Code &

Input

cmd
ums = [0,1,0], k = 1

Output

cmd
2

### #2 Code Example with Python Programming

```Code - Python Programming```

``````
class Solution:
def minKBitFlips(self, a: List[int], k: int) -> int:
q = collections.deque()
res = 0
for i in range(len(a)):
if len(q) % 2 != 0 and a[i] == 1 or len(q) % 2 == a[i] == 0:
res += 1
q.append(i+k-1)
if q and q[0] == i: q.popleft()
if q and q[-1] >= len(a): return -1
return res
``````
Copy The Code &

Input

cmd
ums = [0,1,0], k = 1

Output

cmd
2