# Count of subsets not containing adjacent elements

• Last Updated : 27 May, 2021

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:
All possible subsets are {}, {2} and {7}.

Input: arr[] = {3, 5, 7}
Output: 5

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`

## Javascript

 ``

Output:

`5`

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:

## 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`

## Javascript

 ``

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 method 5 of this article.

My Personal Notes arrow_drop_up