Count of subsets not containing adjacent elements

Given an array arr[] of N integers, the task is to find the count of all the subsets which do not contain adjacent elements from the given array.
Examples:

Input: arr[] = {2, 7} 
Output: 3 
All possible subsets are {}, {2} and {7}.

Input: arr[] = {3, 5, 7} 
Output: 5  

Method 1: Using bit masking 

Idea: The idea is to use a bit-mask pattern to generate all the combinations as discussed in this article. While considering a subset, we need to check if it contains adjacent elements or not. A subset will contain adjacent elements if two or more consecutive bits are set in its bit mask. In order to check if the bit-mask has consecutive bits set or not, we can right shift the mask by one bit and take it AND with the mask. If the result of the AND operation is 0, then the mask does not have consecutive sets and therefore, the corresponding subset will not have adjacent elements from the array.
Below is the implementation of the above approach:  

5

Time Complexity: O(2ⁿ), where n is the size of the given array.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

Method 2: The above approach takes exponential time. In the above code, the number of bit-masks without consecutive 1s was required. This count can be obtained in linear time using dynamic programming as discussed in this article.
Below is the implementation of the above approach:  

5

Time Complexity: O(n)
Auxiliary Space: O(n), where n is the size of the given array

Another approach : Space optimized

 we can space optimize previous approach by using only two variables to store the previous values instead of two arrays. 

Implementation Steps:

• Initialize two variables a and b to 1.
• Use a loop to iterate over the array elements from index 1 to n-1.
• For each element at index i, update a and b using the following formulas:
  a = a + b
  b = previous value of a
• After the loop, compute the total number of possible subsets as the sum of a and b.
• Return the total number of possible subsets.

Implementation:

5

Time Complexity: O(n)
Auxiliary Space: O(1)

Method 3; If we take a closer look at the pattern, we can observe that the count is actually (N + 2)^thFibonacci number for N ? 1.  

n = 1, count = 2 = fib(3) 
n = 2, count = 3 = fib(4) 
n = 3, count = 5 = fib(5) 
n = 4, count = 8 = fib(6) 
n = 5, count = 13 = fib(7) 
................ 

Therefore, the subsets can be counted in O(log n) time using method 5 of this article.