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

难度：Hard

题目：在一个只包含0和1的数组中，每次可以翻转Ｋ个元素，其中所有的0翻转成1,1翻转成0. 返回其最小的翻转次数。如果不存在则返回－1.

思路：贪心算法，每次找到第1个0后翻转从该元素开始的Ｋ个元素。

Runtime: 140 ms, faster than 71.57% of Java online submissions for Minimum Number of K Consecutive Bit Flips.
Memory Usage: 48.4 MB, less than 6.43% of Java online submissions for Minimum Number of K Consecutive Bit Flips.

class Solution {
    public int minKBitFlips(int[] A, int K) {
        int i = 0, count = 0;
        while (i <= A.length - 1) {
            if (1 == A[i]) {
                i++;
                continue;
            }
            if (i + K > A.length) {
                return -1;
            }
            count++;
            for (int t = i; t < i + K; t++) {
                A[t] = (A[t] + 1) & 1;
            }
        }
        
        return count;
    }
}