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 <= A.length <= 300001 <= K <= A.length
在仅包含 0 和 1 的数组 A 中，一次 K 位翻转包括选择一个长度为 K的（连续）子数组，同时将子数组中的每个 0 更改为 1，而每个 1 更改为 0。
返回所需的 K 位翻转的次数，以便数组没有值为 0 的元素。如果不可能，返回 -1。
示例 1：
输入：A = [0,1,0], K = 1 输出：2 解释：先翻转 A[0]，然后翻转 A[2]。
示例 2：
输入：A = [1,1,0], K = 2 输出：-1 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]。
示例 3：
输入：A = [0,0,0,1,0,1,1,0], K = 3 输出：3 解释： 翻转 A[0],A[1],A[2]: A变成 [1,1,1,1,0,1,1,0] 翻转 A[4],A[5],A[6]: A变成 [1,1,1,1,1,0,0,0] 翻转 A[5],A[6],A[7]: A变成 [1,1,1,1,1,1,1,1]
提示：
1 <= A.length <= 300001 <= K <= A.length
Runtime: 752 ms
Memory Usage: 19.1 MB
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 //异或 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 }