# Count sub-sets that satisfy the given condition

Given an array arr[] and an integer x, the task is to count the number of sub-sets of arr[] sum of all of whose sub-sets (individually) is divisible by x.

Examples:

Input: arr[] = {2, 4, 3, 7}, x = 2
Output: 3
All valid sub-sets are {2}, {4} and {2, 4}
{2} => 2 is divisible by 2
{4} => 4 is divisible by 2
{2, 4} => 2, 4 and 6 are all divisible by 2

Input: arr[] = {2, 3, 4, 5}, x = 1
Output: 15

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

Approach: To choose a sub-set sum of all of whose sub-sets is divisible by x, all the elements of the sub-set must be divisible by x. So,

• Count all the element from the array that are divisible by x and store it in a variable count.
• Now, all possible sub-sets satisfying the condition will be 2count – 1

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `#define ll long long int ` `using` `namespace` `std; ` ` `  `// Function to return the count of the required sub-sets ` `ll count(``int` `arr[], ``int` `n, ``int` `x) ` `{ ` ` `  `    ``// Every element is divisible by 1 ` `    ``if` `(x == 1) { ` `        ``ll ans = ``pow``(2, n) - 1; ` `        ``return` `ans; ` `    ``} ` ` `  `    ``// Count of elements which are divisible by x ` `    ``int` `count = 0; ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``if` `(arr[i] % x == 0) ` `            ``count++; ` `    ``} ` ` `  `    ``ll ans = ``pow``(2, count) - 1; ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 2, 4, 3, 5 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``int` `x = 1; ` `    ``cout << count(arr, n, x) << endl; ` `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` ` `  `class` `solution ` `{ ` ` `  `// Function to return the count of the required sub-sets ` `static` `long` `count(``int` `arr[], ``int` `n, ``int` `x) ` `{ ` ` `  `    ``// Every element is divisible by 1 ` `    ``if` `(x == ``1``) { ` `        ``long` `ans = (``long``)Math.pow(``2``, n) - ``1``; ` `        ``return` `ans; ` `    ``} ` ` `  `    ``// Count of elements which are divisible by x ` `    ``int` `count = ``0``; ` `    ``for` `(``int` `i = ``0``; i < n; i++) { ` `        ``if` `(arr[i] % x == ``0``) ` `            ``count++; ` `    ``} ` ` `  `    ``long` `ans = (``long``)Math.pow(``2``, count) - ``1``; ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String args[]) ` `{ ` `    ``int` `[]arr = { ``2``, ``4``, ``3``, ``5` `}; ` `    ``int` `n = arr.length; ` `    ``int` `x = ``1``; ` `    ``System.out.println(count(arr, n, x)); ` `} ` `} `

## Python3

 `# Python3 implementation of the approach  ` ` `  `# Function to return the count of ` `# the required sub-sets  ` `def` `count(arr, n, x) : ` ` `  `    ``# Every element is divisible by 1  ` `    ``if` `(x ``=``=` `1``) : ` `        ``ans ``=` `pow``(``2``, n) ``-` `1` `        ``return` `ans;  ` `     `  `    ``# Count of elements which are  ` `    ``# divisible by x  ` `    ``count ``=` `0` `    ``for` `i ``in` `range``(n) :  ` `        ``if` `(arr[i] ``%` `x ``=``=` `0``) : ` `            ``count ``+``=` `1` ` `  `    ``ans ``=` `pow``(``2``, count) ``-` `1` `    ``return` `ans  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `:  ` ` `  `    ``arr ``=` `[ ``2``, ``4``, ``3``, ``5` `]  ` `    ``n ``=` `len``(arr) ` `    ``x ``=` `1` `    ``print``(count(arr, n, x)) ` ` `  `# This code is contributed by Ryuga `

## C#

 `//C# implementation of the approach ` ` `  `using` `System; ` ` `  `public` `class` `GFG{ ` `     `  `// Function to return the count of the required sub-sets ` `static` `double` `count(``int` `[]arr, ``int` `n, ``int` `x) ` `{ ` `    ``double` `ans=0; ` `    ``// Every element is divisible by 1 ` `    ``if` `(x == 1) { ` `        ``ans = (Math.Pow(2, n) - 1); ` `        ``return` `ans; ` `    ``} ` ` `  `    ``// Count of elements which are divisible by x ` `    ``int` `count = 0; ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``if` `(arr[i] % x == 0) ` `            ``count++; ` `    ``} ` ` `  `    ``ans = (Math.Pow(2, count) - 1); ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `     `  `    ``static` `public` `void` `Main (){ ` `     `  `    ``int` `[]arr = { 2, 4, 3, 5 }; ` `    ``int` `n = arr.Length; ` `    ``int` `x = 1; ` `    ``Console.WriteLine(count(arr, n, x)); ` `    ``} ` `} `

## PHP

 ` `

Output:

```15
```

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