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.
\n\n\n
Example 1:
\n\n\nInput: nums = [0,1,0], k = 1\nOutput: 2\nExplanation: Flip nums[0], then flip nums[2].\n\n\n
Example 2:
\n\n\nInput: nums = [1,1,0], k = 2\nOutput: -1\nExplanation: No matter how we flip subarrays of size 2, we cannot make the array become [1,1,1].\n\n\n
Example 3:
\n\n\nInput: nums = [0,0,0,1,0,1,1,0], k = 3\nOutput: 3\nExplanation: \nFlip nums[0],nums[1],nums[2]: nums becomes [1,1,1,1,0,1,1,0]\nFlip nums[4],nums[5],nums[6]: nums becomes [1,1,1,1,1,0,0,0]\nFlip nums[5],nums[6],nums[7]: nums becomes [1,1,1,1,1,1,1,1]\n\n\n
\n
Constraints:
\n\n1 <= nums.length <= 1051 <= k <= nums.length给定一个二进制数组 nums 和一个整数 k 。
k位翻转 就是从 nums 中选择一个长度为 k 的 子数组 ，同时把子数组中的每一个 0 都改成 1 ，把子数组中的每一个 1 都改成 0 。
返回数组中不存在 0 所需的最小 k位翻转 次数。如果不可能，则返回 -1 。
子数组 是数组的 连续 部分。
\n\n\n\n
示例 1：
\n\n\n输入：nums = [0,1,0], K = 1\n输出：2\n解释：先翻转 A[0]，然后翻转 A[2]。\n\n\n
示例 2：
\n\n\n输入：nums = [1,1,0], K = 2\n输出：-1\n解释：无论我们怎样翻转大小为 2 的子数组，我们都不能使数组变为 [1,1,1]。\n\n\n
示例 3：
\n\n\n输入：nums = [0,0,0,1,0,1,1,0], K = 3\n输出：3\n解释：\n翻转 A[0],A[1],A[2]: A变成 [1,1,1,1,0,1,1,0]\n翻转 A[4],A[5],A[6]: A变成 [1,1,1,1,1,0,0,0]\n翻转 A[5],A[6],A[7]: A变成 [1,1,1,1,1,1,1,1]\n\n\n
\n\n
提示：
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