# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Method 1: 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 its AND with the mask. If the result of the AND operation is 0, then the mask does not have consecutive bits set and therefore, the corresponding subset will not have adjacent elements from the array.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the count ` `// of possible subsets ` `int` `cntSubsets(``int``* arr, ``int` `n) ` `{ ` ` `  `    ``// Total possible subsets of n ` `    ``// sized array is (2^n - 1) ` `    ``unsigned ``int` `max = ``pow``(2, n); ` ` `  `    ``// To store the required ` `    ``// count of subsets ` `    ``int` `result = 0; ` ` `  `    ``// Run from i 000..0 to 111..1 ` `    ``for` `(``int` `i = 0; i < max; i++) { ` `        ``int` `counter = i; ` ` `  `        ``// If current subset has consecutive ` `        ``// elements from the array ` `        ``if` `(counter & (counter >> 1)) ` `            ``continue``; ` `        ``result++; ` `    ``} ` `    ``return` `result; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 3, 5, 7 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` ` `  `    ``cout << cntSubsets(arr, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` `     `  `class` `GFG ` `{ ` ` `  `// Function to return the count ` `// of possible subsets ` `static` `int` `cntSubsets(``int``[] arr, ``int` `n) ` `{ ` ` `  `    ``// Total possible subsets of n ` `    ``// sized array is (2^n - 1) ` `    ``int` `max = (``int``) Math.pow(``2``, n); ` ` `  `    ``// To store the required ` `    ``// count of subsets ` `    ``int` `result = ``0``; ` ` `  `    ``// Run from i 000..0 to 111..1 ` `    ``for` `(``int` `i = ``0``; i < max; i++)  ` `    ``{ ` `        ``int` `counter = i; ` ` `  `        ``// If current subset has consecutive ` `        ``if` `((counter & (counter >> ``1``)) > ``0``) ` `            ``continue``; ` `        ``result++; ` `    ``} ` `    ``return` `result; ` `} ` ` `  `// Driver code ` `static` `public` `void` `main (String []arg) ` `{ ` `    ``int` `arr[] = { ``3``, ``5``, ``7` `}; ` `    ``int` `n = arr.length; ` ` `  `    ``System.out.println(cntSubsets(arr, n)); ` `} ` `} ` ` `  `// This code is contributed by Rajput-Ji `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function to return the count ` `# of possible subsets ` `def` `cntSubsets(arr, n): ` ` `  `    ``# Total possible subsets of n ` `    ``# sized array is (2^n - 1) ` `    ``max` `=` `pow``(``2``, n) ` ` `  `    ``# To store the required ` `    ``# count of subsets ` `    ``result ``=` `0` ` `  `    ``# Run from i 000..0 to 111..1 ` `    ``for` `i ``in` `range``(``max``): ` `        ``counter ``=` `i ` ` `  `        ``# If current subset has consecutive ` `        ``# elements from the array ` `        ``if` `(counter & (counter >> ``1``)): ` `            ``continue` `        ``result ``+``=` `1` ` `  `    ``return` `result ` ` `  `# Driver code ` `arr ``=` `[``3``, ``5``, ``7``] ` `n ``=` `len``(arr) ` ` `  `print``(cntSubsets(arr, n)) ` ` `  `# This code is contributed by Mohit Kumar `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `// Function to return the count ` `// of possible subsets ` `static` `int` `cntSubsets(``int``[] arr, ``int` `n) ` `{ ` ` `  `    ``// Total possible subsets of n ` `    ``// sized array is (2^n - 1) ` `    ``int` `max = (``int``) Math.Pow(2, n); ` ` `  `    ``// To store the required ` `    ``// count of subsets ` `    ``int` `result = 0; ` ` `  `    ``// Run from i 000..0 to 111..1 ` `    ``for` `(``int` `i = 0; i < max; i++)  ` `    ``{ ` `        ``int` `counter = i; ` ` `  `        ``// If current subset has consecutive ` `        ``if` `((counter & (counter >> 1)) > 0) ` `            ``continue``; ` `        ``result++; ` `    ``} ` `    ``return` `result; ` `} ` ` `  `// Driver code ` `static` `public` `void` `Main (String []arg) ` `{ ` `    ``int` `[]arr = { 3, 5, 7 }; ` `    ``int` `n = arr.Length; ` ` `  `    ``Console.WriteLine(cntSubsets(arr, n)); ` `} ` `} ` `     `  `// This code is contributed by Rajput-Ji `

Output:

```5
```

Method 2: The above approach takes exponential time. In the above code, the number of bit-masks without consecutive 1s were 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:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the count ` `// of possible subsets ` `int` `cntSubsets(``int``* arr, ``int` `n) ` `{ ` `    ``int` `a[n], b[n]; ` ` `  `    ``a[0] = b[0] = 1; ` ` `  `    ``for` `(``int` `i = 1; i < n; i++) { ` ` `  `        ``// If previous element was 0 then 0 ` `        ``// as well as 1 can be appended ` `        ``a[i] = a[i - 1] + b[i - 1]; ` ` `  `        ``// If previous element was 1 then ` `        ``// only 0 can be appended ` `        ``b[i] = a[i - 1]; ` `    ``} ` ` `  `    ``// Store the count of all possible subsets ` `    ``int` `result = a[n - 1] + b[n - 1]; ` ` `  `    ``return` `result; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 3, 5, 7 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` ` `  `    ``cout << cntSubsets(arr, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` ` `  `class` `GFG  ` `{ ` ` `  `// Function to return the count ` `// of possible subsets ` `static` `int` `cntSubsets(``int` `[]arr, ``int` `n) ` `{ ` `    ``int` `[]a = ``new` `int``[n]; ` `    ``int` `[]b = ``new` `int``[n]; ` ` `  `    ``a[``0``] = b[``0``] = ``1``; ` ` `  `    ``for` `(``int` `i = ``1``; i < n; i++)  ` `    ``{ ` ` `  `        ``// If previous element was 0 then 0 ` `        ``// as well as 1 can be appended ` `        ``a[i] = a[i - ``1``] + b[i - ``1``]; ` ` `  `        ``// If previous element was 1 then ` `        ``// only 0 can be appended ` `        ``b[i] = a[i - ``1``]; ` `    ``} ` ` `  `    ``// Store the count of all possible subsets ` `    ``int` `result = a[n - ``1``] + b[n - ``1``]; ` ` `  `    ``return` `result; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `arr[] = { ``3``, ``5``, ``7` `}; ` `    ``int` `n = arr.length; ` ` `  `    ``System.out.println(cntSubsets(arr, n)); ` `} ` `} ` ` `  `// This code is contributed by Princi Singh `

## Python3

 `# Python3 implementation of the approach  ` ` `  `# Function to return the count  ` `# of possible subsets  ` `def` `cntSubsets(arr, n) :  ` ` `  `    ``a ``=` `[``0``] ``*` `n ` `    ``b ``=` `[``0``] ``*` `n;  ` ` `  `    ``a[``0``] ``=` `b[``0``] ``=` `1``;  ` ` `  `    ``for` `i ``in` `range``(``1``, n) : ` `         `  `        ``# If previous element was 0 then 0  ` `        ``# as well as 1 can be appended  ` `        ``a[i] ``=` `a[i ``-` `1``] ``+` `b[i ``-` `1``];  ` ` `  `        ``# If previous element was 1 then  ` `        ``# only 0 can be appended  ` `        ``b[i] ``=` `a[i ``-` `1``];  ` ` `  `    ``# Store the count of all possible subsets  ` `    ``result ``=` `a[n ``-` `1``] ``+` `b[n ``-` `1``];  ` ` `  `    ``return` `result;  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `:  ` ` `  `    ``arr ``=` `[ ``3``, ``5``, ``7` `];  ` `    ``n ``=` `len``(arr);  ` ` `  `    ``print``(cntSubsets(arr, n));  ` ` `  `# This code is contributed by AnkitRai01 `

## C#

 `// C# implementation of the approach ` `using` `System; ` `     `  `class` `GFG  ` `{ ` ` `  `// Function to return the count ` `// of possible subsets ` `static` `int` `cntSubsets(``int` `[]arr, ``int` `n) ` `{ ` `    ``int` `[]a = ``new` `int``[n]; ` `    ``int` `[]b = ``new` `int``[n]; ` ` `  `    ``a[0] = b[0] = 1; ` ` `  `    ``for` `(``int` `i = 1; i < n; i++)  ` `    ``{ ` ` `  `        ``// If previous element was 0 then 0 ` `        ``// as well as 1 can be appended ` `        ``a[i] = a[i - 1] + b[i - 1]; ` ` `  `        ``// If previous element was 1 then ` `        ``// only 0 can be appended ` `        ``b[i] = a[i - 1]; ` `    ``} ` ` `  `    ``// Store the count of all possible subsets ` `    ``int` `result = a[n - 1] + b[n - 1]; ` ` `  `    ``return` `result; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `[]arr = { 3, 5, 7 }; ` `    ``int` `n = arr.Length; ` ` `  `    ``Console.WriteLine(cntSubsets(arr, n)); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:

```5
```

Method 3; If we take a closer look at the pattern, we can observe that the count is actually (N + 2)th Fibonacci 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 the method 5 of this article.

My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.