# 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

**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
**2**^{count}– 1

Below is the implementation of the above approach:

## C++

`// C++ implementation of the approach ` `#include <bits/stdc++.h> ` `#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 divisble 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[0]); ` ` ` `int` `x = 1; ` ` ` `cout << count(arr, n, x) << endl; ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

## 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 divisble 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)); ` `} ` `} ` |

*chevron_right*

*filter_none*

## Python3

`# Python3 implementation of the approach ` ` ` `# Function to return the count of ` `# the required sub-sets ` `def` `count(arr, n, x) : ` ` ` ` ` `# Every element is divisble 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 ` |

*chevron_right*

*filter_none*

## 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 divisble 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)); ` ` ` `} ` `} ` |

*chevron_right*

*filter_none*

## PHP

`<?php ` `// PHP implementation of the approach ` ` ` `// Function to return the count of the required sub-sets ` `function` `count_t(` `$arr` `, ` `$n` `, ` `$x` `) ` `{ ` ` ` `// Every element is divisble by 1 ` ` ` `if` `(` `$x` `== 1) { ` ` ` `$ans` `= pow(2, ` `$n` `) - 1; ` ` ` `return` `$ans` `; ` ` ` `} ` ` ` ` ` `// Count of elements which are divisible by x ` ` ` `$count` `= 0; ` ` ` `for` `(` `$i` `= 0; ` `$i` `< ` `$n` `; ` `$i` `++) { ` ` ` `if` `(` `$arr` `[` `$i` `] % ` `$x` `== 0) ` ` ` `$count` `++; ` ` ` `} ` ` ` ` ` `$ans` `= pow(2, ` `$count` `) - 1; ` ` ` `return` `$ans` `; ` `} ` ` ` `// Driver code ` ` ` ` ` `$arr` `= ` `array` `( 2, 4, 3, 5 ); ` ` ` `$n` `= sizeof(` `$arr` `) / sizeof(` `$arr` `[0]); ` ` ` `$x` `= 1; ` ` ` `echo` `count_t(` `$arr` `, ` `$n` `, ` `$x` `); ` ` ` `#This code is contributed by akt_mit ` `?> ` |

*chevron_right*

*filter_none*

**Output:**

15

## Recommended Posts:

- Count index pairs which satisfy the given condition
- Pairs from an array that satisfy the given condition
- Count number of subsets having a particular XOR value
- Count no. of ordered subsets having a particular XOR value
- Count subsets having distinct even numbers
- Count of all possible pairs of disjoint subsets of integers from 1 to N
- Count number of subsets whose median is also present in the same subset
- Count minimum number of subsets (or subsequences) with consecutive numbers
- Partition an array of non-negative integers into two subsets such that average of both the subsets is equal
- Find all pairs (a,b) and (c,d) in array which satisfy ab = cd
- Remove elements to make array satisfy arr[ i+1] < arr[i] for each valid i
- Smallest index in the given array that satisfies the given condition
- Maximum size of sub-array that satisfies the given condition
- Check if elements of an array can be arranged satisfying the given condition
- Divide N segments into two non-empty groups such that given condition is satisfied

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.