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 Kint 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 codeint 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 Kint 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 codeint 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 Kimport java.io.*;class GFG {// Function to find Product of all the elements// in an array divisible by a given number Kstatic 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 Kdef 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 codeif __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 Kusing System;class GFG {// Function to find Product of all // the elements in an array divisible// by a given number Kstatic 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 codepublic 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 Kfunction 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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