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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 \(\textit{nums}\). The space complexity is \(O(1)\).

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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

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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;
    }
}

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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;
    }
};

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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
}

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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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