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 2Input: 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 elements from the array that is divisible by x and store them 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 <bits/stdc++.h>#define ll long long intusing namespace std;// Function to return the count of the required sub-setsll 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 codeint 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;} |
Java
// Java implementation of the approachimport java.util.*;class solution{// Function to return the count of the required sub-setsstatic 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 codepublic 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 approachusing System;public class GFG{ // Function to return the count of the required sub-setsstatic 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
<?php// PHP implementation of the approach// Function to return the count of the required sub-setsfunction count_t($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 = 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?> |
Javascript
<script> // Javascript implementation of the approach // Function to return the count of // the required sub-sets function count(arr, n, x) { let 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 let count = 0; for (let i = 0; i < n; i++) { if (arr[i] % x == 0) count++; } ans = (Math.pow(2, count) - 1); return ans; } let arr = [ 2, 4, 3, 5 ]; let n = arr.length; let x = 1; document.write(count(arr, n, x)); </script> |
Output
15
Complexity Analysis:
- Time complexity: O(n) because using a for loop
- Auxiliary Space: O(1)
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