The solution I got to the problem is below:

public boolean twoTwo(int[] nums) {
    final int length = nums.length;
    for (int i = 0; i < length;){
        int count = 0;      // Used to count following 2's
        while(i < length && nums[i++] == 2){
            count++;
        }
        if(count == 1){ // No adjacent 2's! Set doesn't work.
            return false;
        }
    }
    return true; // Didn't come across a lone 2
}

The way that I handle this, is that I count all the adjacent 2's. If the count is not 1, we are good. This means that there was either no 2 at that index, or a group of 2's was present. This holds, since we traverse the array in a single direction.

A good thing about this solution is that it will work for an array of any size. Note that it would have a linear complexity, even though 2 loops are present. They both just traverse using the same index value, only ever sweeping over the array once.

If at any time we find a 2, then check the following only to find there are 0 following 2's (denoted by count), we return false.