# 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.

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); ` ` `  `    ``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 = b = 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); ` ` `  `    ``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 = b = 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.

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