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3191. Minimum Operations to Make Binary Array Elements Equal to One I

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 <= 105
  • 0 <= nums[i] <= 1

Solutions

Solution 1: Sequential Traversal + Simulation

We notice that the first position in the array that is $0$ must undergo a flip operation, otherwise, it cannot be turned into $1$. Therefore, we can sequentially traverse the array, and each time we encounter $0$, we flip the next two elements and accumulate one operation count.

After the traversal, we return the answer.

The time complexity is $O(n)$, where $n$ is the length of the array $\text{nums}$. The space complexity is $O(1)$.

  • class Solution {
    public int minOperations(int[] nums) {
        int ans = 0;
        int n = nums.length;
        for (int i = 0; i < n; ++i) {
            if (nums[i] == 0) {
                if (i + 2 >= n) {
                    return -1;
                }
                nums[i + 1] ^= 1;
                nums[i + 2] ^= 1;
                ++ans;
            }
        }
        return ans;
    }
}

  • class Solution {
public:
    int minOperations(vector<int>& nums) {
        int ans = 0;
        int n = nums.size();
        for (int i = 0; i < n; ++i) {
            if (nums[i] == 0) {
                if (i + 2 >= n) {
                    return -1;
                }
                nums[i + 1] ^= 1;
                nums[i + 2] ^= 1;
                ++ans;
            }
        }
        return ans;
    }
};

  • class Solution:
    def minOperations(self, nums: List[int]) -> int:
        ans = 0
        for i, x in enumerate(nums):
            if x == 0:
                if i + 2 >= len(nums):
                    return -1
                nums[i + 1] ^= 1
                nums[i + 2] ^= 1
                ans += 1
        return ans


  • func minOperations(nums []int) (ans int) {
	for i, x := range nums {
		if x == 0 {
			if i+2 >= len(nums) {
				return -1
			}
			nums[i+1] ^= 1
			nums[i+2] ^= 1
			ans++
		}
	}
	return
}

  • function minOperations(nums: number[]): number {
    const n = nums.length;
    let ans = 0;
    for (let i = 0; i < n; ++i) {
        if (nums[i] === 0) {
            if (i + 2 >= n) {
                return -1;
            }
            nums[i + 1] ^= 1;
            nums[i + 2] ^= 1;
            ++ans;
        }
    }
    return ans;
}

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