# 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;
}