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Formatted question description: https://leetcode.ca/all/995.html

# 995. Minimum Number of K Consecutive Bit Flips

Hard

## Description

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, then flip A.

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,A,A: A becomes [1,1,1,1,0,1,1,0]

Flip A,A,A: A becomes [1,1,1,1,1,0,0,0]

Flip A,A,A: A becomes [1,1,1,1,1,1,1,1]

Note:

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

## Solution

One straightforward solution is to directly simulate the whole process of flipping. If A is 0, then it must be flipped to 1, so flip the subarray from A to A[K - 1]. After flipping A if necessary, check A, and if A is 0, it must be flipped to 1, so flip the subarray from A to A[K]. Repeat the process such that for each index i from 1 to A.length - K, if A[i] is 0, then flip the subarray A[i] to A[i + K - 1]. After flipping, if there is at least one element in A that is 0, return -1. Otherwise, return the number of flips.

The solution above is not efficient when K is quite large. A better solution is to use sliding window. Use a queue to store tne indices in a sliding window of size K, that is, the difference between the maximum element and the minimum element in the queue is always less than K. For i from 0 to A.length - 1, if the queue’s first element equals i - K, then poll the first element from the queue. Then check whether the subarray that starts from index i and has length K needs to be flipped. This is determined by the number of elements in the queue and the value A[i]. If there are even number of elements in the queue and A[i] is 0, or there are odd number of elements in the queue and A[i] is 1, then the flip must be done at index i. If the flip must be done and i + K > A.length, then the flip is impossible, so return -1. If the flip must be done and i + K <= A.length, then increase the number of flips by 1 and offer i to the queue. Finally, return the number of flips.

• class Solution {
public int minKBitFlips(int[] A, int K) {
int flips = 0;
int length = A.length;
for (int i = 0; i < length; i++) {
if (!queue.isEmpty() && queue.peek() + K == i)
queue.poll();
if (queue.size() % 2 == A[i]) {
if (i + K > length)
return -1;
else {
flips++;
queue.offer(i);
}
}
}
return flips;
}
}

• // OJ: https://leetcode.com/problems/minimum-number-of-k-consecutive-bit-flips/
// Time: O(N)
// Space: O(K)
class Solution {
public:
int minKBitFlips(vector<int>& A, int k) {
queue<int> q;
int ans = 0;
for (int i = 0, N = A.size(); i < N; ++i) {
if (q.size() && q.front() <= i) q.pop();
int flip = q.size() % 2;
if ((A[i] ^ flip) == 0) {
if (i + k > N) return -1;
++ans;
q.push(i + k);
}
}
return ans;
}
};

• # 995. Minimum Number of K Consecutive Bit Flips
# https://leetcode.com/problems/minimum-number-of-k-consecutive-bit-flips/

class Solution:
def minKBitFlips(self, nums: List[int], k: int) -> int:
queue = deque()
res = 0

for i, x in enumerate(nums):
if x == 0:
if len(queue) % 2 == 0:
res += 1
queue.append(i + k - 1)
else:
if len(queue) & 1:
res += 1
queue.append(i + k - 1)

if queue and i >= queue:
queue.popleft()

return res if len(queue) == 0 else -1