Product of all the elements in an array divisible by a given number K

Given an array containing N elements and a number K. The task is to find the product of all such elements of the array which are divisible by K.

Examples:

Input : arr[] = {15, 16, 10, 9, 6, 7, 17}
        K = 3
Output : 810

Input : arr[] = {5, 3, 6, 8, 4, 1, 2, 9}
        K = 2
Output : 384

The idea is to traverse the array and check the elements one by one. If an element is divisible by K then multiply that element’s value with the product so far and continue this process while the end of the array is reached.



Below is the implementation of the above approach:

C++

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// C++ program to find Product of all the elements
// in an array divisible by a given number K
  
#include <iostream>
using namespace std;
  
// Function to find Product of all the elements
// in an array divisible by a given number K
int findProduct(int arr[], int n, int k)
{
    int prod = 1;
  
    // Traverse the array
    for (int i = 0; i < n; i++) {
  
        // If current element is divisible by k
        // multiply with product so far
        if (arr[i] % k == 0) {
            prod *= arr[i];
        }
    }
  
    // Return calculated product
    return prod;
}
  
// Driver code
int main()
{
    int arr[] = { 15, 16, 10, 9, 6, 7, 17 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 3;
  
    cout << findProduct(arr, n, k);
  
    return 0;
}

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Java

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// Java program to find Product of all the elements
// in an array divisible by a given number K
  
import java.io.*;
  
class GFG {
  
// Function to find Product of all the elements
// in an array divisible by a given number K
static int findProduct(int arr[], int n, int k)
{
    int prod = 1;
  
    // Traverse the array
    for (int i = 0; i < n; i++) {
  
        // If current element is divisible by k
        // multiply with product so far
        if (arr[i] % k == 0) {
            prod *= arr[i];
        }
    }
  
    // Return calculated product
    return prod;
}
  
// Driver code
    public static void main (String[] args) {
        int arr[] = { 15, 16, 10, 9, 6, 7, 17 };
    int n = arr.length;
    int k = 3;
  
    System.out.println(findProduct(arr, n, k));
    }
}
  
  
// This code is contributed by inder_verma..

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Python3

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# Python3 program to find Product of all 
# the elements in an array divisible by
# a given number K
  
# Function to find Product of all the elements
# in an array divisible by a given number K
def findProduct(arr, n, k):
  
    prod = 1
  
    # Traverse the array
    for i in range(n):
  
        # If current element is divisible 
        # by k, multiply with product so far
        if (arr[i] % k == 0):
            prod *= arr[i]
  
    # Return calculated product
    return prod
  
# Driver code
if __name__ == "__main__":
  
    arr= [15, 16, 10, 9, 6, 7, 17 ]
    n = len(arr)
    k = 3
  
    print (findProduct(arr, n, k))
  
# This code is contributed by ita_c

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C#

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// C# program to find Product of all 
// the elements in an array divisible
// by a given number K
using System;
  
class GFG 
{
  
// Function to find Product of all 
// the elements in an array divisible
// by a given number K
static int findProduct(int []arr, int n, int k)
{
    int prod = 1;
  
    // Traverse the array
    for (int i = 0; i < n; i++) 
    {
  
        // If current element is divisible 
        // by k multiply with product so far
        if (arr[i] % k == 0) 
        {
            prod *= arr[i];
        }
    }
  
    // Return calculated product
    return prod;
}
  
// Driver code
public static void Main()
{
    int []arr = { 15, 16, 10, 9, 6, 7, 17 };
    int n = arr.Length;
    int k = 3;
      
    Console.WriteLine(findProduct(arr, n, k));
}
}
  
// This code is contributed by inder_verma

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PHP

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<?php
// PHP program to find Product of 
// all the elements in an array 
// divisible by a given number K 
  
// Function to find Product of 
// all the elements in an array 
// divisible by a given number K 
function findProduct(&$arr, $n, $k
    $prod = 1; 
  
    // Traverse the array 
    for ($i = 0; $i < $n; $i++) 
    
  
        // If current element is divisible  
        // by k multiply with product so far 
        if ($arr[$i] % $k == 0) 
        
            $prod *= $arr[$i]; 
        
    
  
    // Return calculated product 
    return $prod
  
// Driver code 
$arr = array(15, 16, 10, 9, 6, 7, 17 ); 
$n = sizeof($arr); 
$k = 3; 
  
echo (findProduct($arr, $n, $k)); 
  
// This code is contributed
// by Shivi_Aggarwal
?>

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Output:

810

Time Complexity: O(N), where N is the number of elements in the array.



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