Given an array arr[] of N positive integers. The task is to find the sum of all composite elements which are divisible by a given number k in the given array.
Examples:
Input: arr[] = {1, 3, 4, 5, 7}, k = 2
Output: 4, 4
There is one composite number i.e. 4.
So, sum = 4 and product = 4
Input: arr[] = {1, 2, 3, 4, 5, 6, 7}, k = 2
Output: 10, 24
There is only two composite numbers i.e. 4 and 6.
So, sum = 10 and product = 24
Naive Approach: A simple solution is to traverse the array and keep checking for every element if it is composite or not and also divisible by k then add and product the composite element at the same time.
Efficient Approach: Generate all primes up to the maximum element of the array using the sieve of Eratosthenes and store them in a hash. Now traverse the array and find the sum and product of those elements which are composite and divisible by k using the sieve.
Below is the implementation of the efficient approach:
C++
// CPP program to find sum and product of // composites which are divisible by k in given array.#include <bits/stdc++.h>using namespace std;// Function to find count of compositevoid compositeSumProduct(int arr[], int n, int k){ // Find maximum value in the array int max_val = *max_element(arr, arr + n); // Use sieve to find all prime numbers less than // or equal to max_val // Create a boolean array "prime[0..n]". A // value in prime[i] will finally be false // if i is Not a prime, else true. vector<bool> prime(max_val + 1, true); // Remaining part of SIEVE prime[0] = true; prime[1] = true; for (int p = 2; p * p <= max_val; p++) { // If prime[p] is not changed, then // it is a prime if (prime[p] == true) { // Update all multiples of p for (int i = p * 2; i <= max_val; i += p) prime[i] = false; } } // Sum all primes in arr[] int sum = 0, product = 1; for (int i = 0; i < n; i++) if (!prime[arr[i]] && arr[i] % k == 0) { sum += arr[i]; product *= arr[i]; } cout << "Sum of composite numbers divisible by k is " << sum; cout << "\nProduct of composite numbers divisible by k is " << product;}// Driver codeint main(){ int arr[] = { 1, 2, 3, 4, 5, 6, 7 }; int n = sizeof(arr) / sizeof(arr[0]); int k = 2; compositeSumProduct(arr, n, k); return 0;} |
Java
// Java program to find sum and product of // composites which are divisible by k in given array. import java.util.Vector;public class GFG { static int max_val(int[] arr) { int i; // Initialize maximum element int max = arr[0]; // Traverse array elements from second and // compare every element with current max for (i = 1; i < arr.length; i++) { if (arr[i] > max) { max = arr[i]; } } return max; }// Function to find count of composite static void compositeSumProduct(int arr[], int n, int k) { // Find maximum value in the array int max_val = max_val(arr); // Use sieve to find all prime numbers less than // or equal to max_val // Create a boolean array "prime[0..n]". A // value in prime[i] will finally be false // if i is Not a prime, else true. Vector<Boolean> prime = new Vector<>(); // Remaining part of SIEVE for(int i = 0;i<max_val+1;i++){ prime.add(i, Boolean.TRUE); } for (int p = 2; p * p <= max_val; p++) { // If prime[p] is not changed, then // it is a prime if (prime.get(p) == true) { // Update all multiples of p for (int i = p * 2; i <= max_val; i += p) { prime.add(i, Boolean.FALSE); } } } // Sum all primes in arr[] int sum = 0, product = 1; for (int i = 0; i < n; i++) { //if (!prime[arr[i]] && arr[i] % k == 0) if (!prime.get(arr[i]) && arr[i] % k == 0) { sum += arr[i]; product *= arr[i]; } } System.out.println("Sum of composite numbers " + "divisible by k is " + sum); System.out.println("\nProduct of composite numbers" + " divisible by k is " + product); }// Driver code public static void main(String[] args) { int arr[] = {1, 2, 3, 4, 5, 6, 7}; int n = arr.length; int k = 2; compositeSumProduct(arr, n, k); }} |
Python3
# Python 3 program to find sum and product # of composites which are divisible by k # in given array.from math import sqrt, ceil# Function to find count of compositedef compositeSumProduct(arr, n, k): # Find maximum value in the array max_val = arr[0]; for i in range(len(arr)): if(max_val < arr[i]): max_val = arr[i] # Use sieve to find all prime numbers # less than or equal to max_val # Create a boolean array "prime[0..n]". # A value in prime[i] will finally be # false if i is Not a prime, else true. prime = [True for i in range(max_val + 1)] # Remaining part of SIEVE k = int(sqrt(max_val)) for p in range(2, k + 1, 1): # If prime[p] is not changed, # then it is a prime if (prime[p] == True): # Update all multiples of p for i in range(p * 2, max_val, p): prime[i] = False # Sum all primes in arr[] sum = 0 product = 1 for i in range(0, n, 1): if (prime[arr[i]] == False and arr[i] % k == 0): sum += arr[i] product *= arr[i] print("Sum of composite numbers", "divisible by k is", sum) print("Product of composite numbers", "divisible by k is", product)# Driver codeif __name__ == '__main__': arr = [1, 2, 3, 4, 5, 6, 7] n = len(arr) k = 2 compositeSumProduct(arr, n, k)# This code is contributed by# Surendra_Gangwar |
C#
// C# program to find sum and product of // composites which are divisible by k in given array. using System; using System.Collections.Generic;public class GFG { static int max_val_func(int []arr) { int i; // Initialize maximum element int max = arr[0]; // Traverse array elements from second and // compare every element with current max for (i = 1; i < arr.Length; i++) { if (arr[i] > max) { max = arr[i]; } } return max; } // Function to find count of composite static void compositeSumProduct(int []arr, int n, int k) { // Find maximum value in the array int max_val = max_val_func(arr); // Use sieve to find all prime numbers less than // or equal to max_val // Create a boolean array "prime[0..n]". A // value in prime[i] will finally be false // if i is Not a prime, else true. List<int> prime = new List<int>(); // Remaining part of SIEVE for(int i = 0;i<max_val+1;i++){ prime.Insert(i, 1); } for (int p = 2; p * p <= max_val; p++) { // If prime[p] is not changed, then // it is a prime if (prime[p] == 1) { // Update all multiples of p for (int i = p * 2; i <= max_val; i += p) { prime.Insert(i, 0); } } } // Sum all primes in arr[] int sum = 0, product = 1; for (int i = 0; i < n; i++) { //if (!prime[arr[i]] && arr[i] % k == 0) if (prime[arr[i]]==0 && arr[i] % k == 0) { sum += arr[i]; product *= arr[i]; } } Console.WriteLine("Sum of composite numbers " + "divisible by k is " + sum); Console.WriteLine("\nProduct of composite numbers" + " divisible by k is " + product); } // Driver code static public void Main(String []args) { int []arr = {1, 2, 3, 4, 5, 6, 7}; int n = arr.Length; int k = 2; compositeSumProduct(arr, n, k); } } //contributed by Arnab Kundu |
Javascript
<script>// Javascript program to find sum and product of// composites which are divisible by k in given array.// Function to find count of compositefunction compositeSumProduct(arr, n, k) { // Find maximum value in the array let max_val = arr.sort((a, b) => b - a)[0]; // Use sieve to find all prime numbers less than // or equal to max_val // Create a boolean array "prime[0..n]". A // value in prime[i] will finally be false // if i is Not a prime, else true. let prime = new Array(max_val + 1).fill(true); // Remaining part of SIEVE prime[0] = true; prime[1] = true; for (let p = 2; p * p <= max_val; p++) { // If prime[p] is not changed, then // it is a prime if (prime[p] == true) { // Update all multiples of p for (let i = p * 2; i <= max_val; i += p) prime[i] = false; } } // Sum all primes in arr[] let sum = 0, product = 1; for (let i = 0; i < n; i++) if (!prime[arr[i]] && arr[i] % k == 0) { sum += arr[i]; product *= arr[i]; } document.write("Sum of composite numbers divisible by k is " + sum); document.write("<br>Product of composite numbers divisible by k is " + product);}// Driver codelet arr = [1, 2, 3, 4, 5, 6, 7];let n = arr.lengthlet k = 2;compositeSumProduct(arr, n, k);// This code is contributed by gfgking</script> |
Output
Sum of composite numbers divisible by k is 10 Product of composite numbers divisible by k is 24
Complexity Analysis:
- Time complexity: O(n), as we are looping over n times. Where n is the maximum value of an element in an array, not size.
- Auxiliary Space: O(MaxElement), as we are using a prime array of size MaxElement.
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