In an array A containing only 0s and 1s, a K-bit flip consists of choosing a (contiguous) subarray of length K 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

Example 1:

Input: A = [0,1,0], K = 1
Output: 2
Explanation: Flip A[0], then flip A[2].

Example 2:

Input: A = [1,1,0], K = 2
Output: -1
Explanation: No matter how we flip subarrays of size 2, we can‘t make the array become [1,1,1].

Example 3:

Input: A = [0,0,0,1,0,1,1,0], K = 3
Output: 3
Explanation:
Flip A[0],A[1],A[2]: A becomes [1,1,1,1,0,1,1,0]
Flip A[4],A[5],A[6]: A becomes [1,1,1,1,1,0,0,0]
Flip A[5],A[6],A[7]: A becomes [1,1,1,1,1,1,1,1]
 
 

Note:

1. 1 <= A.length <= 30000
2. 1 <= K <= A.length

在僅包含 0 和 1 的數組 A 中，一次 K 位翻轉包括選擇一個長度為 K的（連續）子數組，同時將子數組中的每個 0 更改為 1，而每個 1 更改為 0。

返回所需的 K 位翻轉的次數，以便數組沒有值為 0 的元素。如果不可能，返回 -1。

示例 1：

輸入：A = [0,1,0], K = 1
輸出：2
解釋：先翻轉 A[0]，然後翻轉 A[2]。

示例 2：

輸入：A = [1,1,0], K = 2
輸出：-1
解釋：無論我們怎樣翻轉大小為 2 的子數組，我們都不能使數組變為 [1,1,1]。

示例 3：

輸入：A = [0,0,0,1,0,1,1,0], K = 3
輸出：3
解釋：
翻轉 A[0],A[1],A[2]: A變成 [1,1,1,1,0,1,1,0]
翻轉 A[4],A[5],A[6]: A變成 [1,1,1,1,1,0,0,0]
翻轉 A[5],A[6],A[7]: A變成 [1,1,1,1,1,1,1,1] 

提示：

1. 1 <= A.length <= 30000
2. 1 <= K <= A.length

 1 class Solution {
 2     func minKBitFlips(_ A: [Int], _ K: Int) -> Int {
 3         var ans:Int = 0
 4         var n:Int = A.count
 5         var f:[Bool] = [Bool](repeating:false,count:n + 1)
 6         var sum:Bool = false
 7         for i in 0..<n
 8         {
 9             //異或
10             sum = (sum == f[i]) ? false : true
11             if sum == (A[i] == 1)
12             {
13                 if i + K > n {return -1}
14                 f[i + K] = !f[i + K]
15                 sum = !sum
16                 ans += 1
17             }
18         }
19         return ans        
20     }
21 }

[Swift Weekly Contest 124]LeetCode995. K 連續位的最小翻轉次數 | Minimum Number of K Consecutive Bit Flips