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995. Minimum Number of K Consecutive Bit Flips

Difficulty Topics

Description

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.

 

Example 1:

Input: nums = [0,1,0], k = 1
Output: 2
Explanation: Flip nums[0], then flip nums[2].

Example 2:

Input: nums = [1,1,0], k = 2
Output: -1
Explanation: No matter how we flip subarrays of size 2, we cannot make the array become [1,1,1].

Example 3:

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

 

Constraints:

  • 1 <= nums.length <= 105
  • 1 <= k <= nums.length

Solution

minimum-number-of-k-consecutive-bit-flips.py
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[0]:
                queue.popleft()

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