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++
// 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;
} |
C
// C program to find Product of all the elements // in an array divisible by a given number K #include <stdio.h> // 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;
printf ( "%d" ,findProduct(arr, n, k));
return 0;
} // This code is contributed by kothavvsaakash. |
Java
// 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.. |
Python3
# 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 |
C#
// 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 |
PHP
<?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 ?> |
Javascript
<script> // Function to find Product of all the elements // in an array divisible by a given number K function findProduct( arr, n, k)
{ var prod = 1;
// Traverse the array
for ( var 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;
} var arr = [15, 16, 10, 9, 6, 7, 17 ];
document.write(findProduct(arr, 7, 3));
</script> |
Output
810
Complexity Analysis:
-
Time Complexity: O(N), where N is the number of elements in the array.
Auxiliary Space: O(1)