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

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

Solution 1: Sequential Traversal + Simulation

JAVA

• 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;
      }
  }

C++

• 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;
      }
  };