Minimum Number of K Consecutive Bit Flips
<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>
Bit Manipulation; Queue; Array; Prefix Sum; Sliding Window
class Solution {
    public int minKBitFlips(int[] nums, int k) {
        int n = nums.length;
        int[] d = new int[n + 1];
        int ans = 0, s = 0;
        for (int i = 0; i < n; ++i) {
            s += d[i];
            if (s % 2 == nums[i]) {
                if (i + k > n) {
                    return -1;
                }
                ++d[i];
                --d[i + k];
                ++s;
                ++ans;
            }
        }
        return ans;
    }
}
