<p>You are given a binary array <code>nums</code> and an integer <code>k</code>.</p>

<p>A <strong>k-bit flip</strong> is choosing a <strong>subarray</strong> of length <code>k</code> from <code>nums</code> and simultaneously changing every <code>0</code> in the subarray to <code>1</code>, and every <code>1</code> in the subarray to <code>0</code>.</p>

<p>Return <em>the minimum number of <strong>k-bit flips</strong> required so that there is no </em><code>0</code><em> in the array</em>. If it is not possible, return <code>-1</code>.</p>

<p>A <strong>subarray</strong> is a <strong>contiguous</strong> part of an array.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [0,1,0], k = 1
<strong>Output:</strong> 2
<strong>Explanation:</strong> Flip nums[0], then flip nums[2].
</pre>

<p><strong class="example">Example 2:</strong></p>

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

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [0,0,0,1,0,1,1,0], k = 3
<strong>Output:</strong> 3
<strong>Explanation:</strong> 
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]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= k &lt;= nums.length</code></li>
</ul>

class Solution:
    def minKBitFlips(self, nums: List[int], k: int) -> int:
        next_inversion = deque()
        flips = 0
        target = 1
        for i in range(len(nums)):
            if next_inversion and i == next_inversion[0]:
                target = 0 if target == 1 else 1
                next_inversion.popleft()
            if nums[i] == target:
                continue
            if i + k > len(nums):
                flips = -1
                break
            flips += 1
            target = 0 if target == 1 else 1
            next_inversion.append(i + k)
        
        return flips

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