Find the count of subsequences where each element is divisible by K

Given an array arr[] and an integer K, the task is to find the total number of subsequences from the array where each element is divisible by K.

Examples:

Input: arr[] = {1, 2, 3, 6}, K = 3
Output: 3
{3}, {6} and {3, 6} are the only valid subsequences.

Input: arr[] = {5, 10, 15, 20, 25}, K = 5
Output: 31

Approach: Since each of the element must be divisible by K, total subsequences are equal to 2cnt where cnt is the number of elements in the array that are divisible by K. Note that 1 will be subtracted from the result in order to exclude the empty subsequence. So, the final result will be 2cnt – 1.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the count
// of all valid subsequences
int countSubSeq(int arr[], int n, int k)
{
  
    // To store the count of elements
    // which are divisible by k
    int count = 0;
  
    for (int i = 0; i < n; i++) {
  
        // If current element is divisible by
        // k then increment the count
        if (arr[i] % k == 0) {
            count++;
        }
    }
  
    // Total (2^n - 1) non-empty subsequences
    // are possible with n element
    return (pow(2, count) - 1);
}
  
// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 3;
  
    cout << countSubSeq(arr, n, k);
  
    return 0;
}

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Java

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// Java implementation of the approach
import java.util.*;
class GFG 
{
  
// Function to return the count
// of all valid subsequences
static int countSubSeq(int arr[], int n, int k)
{
  
    // To store the count of elements
    // which are divisible by k
    int count = 0;
  
    for (int i = 0; i < n; i++)
    {
  
        // If current element is divisible by
        // k then increment the count
        if (arr[i] % k == 0
        {
            count++;
        }
    }
  
    // Total (2^n - 1) non-empty subsequences
    // are possible with n element
    return (int) (Math.pow(2, count) - 1);
}
  
// Driver code
public static void main(String[] args) 
{
    int arr[] = { 1, 2, 3, 6 };
    int n = arr.length;
    int k = 3;
  
    System.out.println(countSubSeq(arr, n, k));
}
  
// This code is contributed by Rajput-Ji

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Python3

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# Python3 implementation of the approach 
  
# Function to return the count 
# of all valid subsequences 
def countSubSeq(arr, n, k) :
  
    # To store the count of elements 
    # which are divisible by k 
    count = 0
  
    for i in range(n) : 
  
        # If current element is divisible by 
        # k then increment the count 
        if (arr[i] % k == 0) :
            count += 1
  
    # Total (2^n - 1) non-empty subsequences 
    # are possible with n element 
    return (2 ** count - 1); 
  
# Driver code 
if __name__ == "__main__"
  
    arr = [ 1, 2, 3, 6 ]; 
    n = len(arr); 
    k = 3
  
    print(countSubSeq(arr, n, k)); 
  
# This code is contributed by AnkitRai01

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C#

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// C# implementation of the approach
using System;
      
class GFG 
{
  
// Function to return the count
// of all valid subsequences
static int countSubSeq(int []arr, int n, int k)
{
  
    // To store the count of elements
    // which are divisible by k
    int count = 0;
  
    for (int i = 0; i < n; i++)
    {
  
        // If current element is divisible by
        // k then increment the count
        if (arr[i] % k == 0) 
        {
            count++;
        }
    }
  
    // Total (2^n - 1) non-empty subsequences
    // are possible with n element
    return (int) (Math.Pow(2, count) - 1);
}
  
// Driver code
public static void Main(String[] args) 
{
    int []arr = { 1, 2, 3, 6 };
    int n = arr.Length;
    int k = 3;
  
    Console.WriteLine(countSubSeq(arr, n, k));
}
  
// This code is contributed by 29AjayKumar

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Output:

3


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