<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:
        n= len(nums)
        flipCountOffset= [0]* n
        flips = 0
        ans = 0
        
        for i in range(n):
            flips+= flipCountOffset[i]
            num = nums[i]^1 if flips % 2 else nums[i]
            nums[i] =  num
            # print(nums)
            
            if num == 0:
                if i+k > n:
                    return -1
                
                nums[i] ^= 1
                if i+k < n:
                    flipCountOffset[i+k]-=1
                flips+=1
                ans+=1 
                
        return ans
19800