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)
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