Description
You are given a binary array nums.
You can do the following operation on the array any number of times (possibly zero):
- Choose any 3 consecutive elements from the array and flip all of them.
Flipping an element means changing its value from 0 to 1, and from 1 to 0.
Return the minimum number of operations required to make all elements in nums equal to 1. If it is impossible, return -1.
Example 1:
Input: nums = [0,1,1,1,0,0]
Output: 3
Explanation:
We can do the following operations:
- Choose the elements at indices 0, 1 and 2. The resulting array is
nums = [1,0,0,1,0,0]. - Choose the elements at indices 1, 2 and 3. The resulting array is
nums = [1,1,1,0,0,0]. - Choose the elements at indices 3, 4 and 5. The resulting array is
nums = [1,1,1,1,1,1].
Example 2:
Input: nums = [0,1,1,1]
Output: -1
Explanation:
It is impossible to make all elements equal to 1.
Constraints:
3 <= nums.length <= 1050 <= nums[i] <= 1
Solution
Python3
class Solution:
def minOperations(self, nums: List[int]) -> int:
N = len(nums)
count = [0] * (N + 1)
res = 0
for i in range(N):
if i - 1 >= 0:
count[i] += count[i - 1]
if i < N - 2 and (nums[i] + count[i]) % 2 == 0:
count[i] += 1
count[i + 3] -= 1
res += 1
if (nums[i] + count[i]) % 2 == 0:
return -1
return res