LeetCode-in-Java
3191. Minimum Operations to Make Binary Array Elements Equal to One I
Medium
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
public class Solution {
public int minOperations(int[] nums) {
int ans = 0;
// Iterate through the array up to the third-last element
for (int i = 0; i < nums.length - 2; i++) {
// If the current element is 0, perform an operation
if (nums[i] == 0) {
ans++;
// Flip the current element and the next two elements
nums[i] = 1;
nums[i + 1] = nums[i + 1] == 0 ? 1 : 0;
nums[i + 2] = nums[i + 2] == 0 ? 1 : 0;
}
}
// Check the last two elements if they are 0, return -1 as they cannot be flipped
for (int i = nums.length - 2; i < nums.length; i++) {
if (nums[i] == 0) {
return -1;
}
}
return ans;
}
}