Given two arrays arr1[] and arr2[]. The task is to find the count of such elements in the first array whose at-least one factor is present in the second array.
Examples:
Input : arr1[] = {10, 2, 13, 4, 15} ; arr2[] = {2, 4, 5, 6}
Output : 4
There is no factor of 13 which is present in the
second array. Except 13, factors of the rest 4
elements of the first array is present in the
second array.
Input : arr1[] = {11, 13, 17, 15} ; arr2[] = {3, 7, 9, 5}
Output : 1
The idea is to insert all elements of the second array into a hash such that the lookup for factors can be done in constant time. Now, traverse the first array and for every element generate all of the factors starting from 1 and check if any of the factors is present in the hash or not.
Below is the implementation of the above approach:
C++
// CPP program to find count of// elements in first array whose// atleast one factor is present// in second array.#include <bits/stdc++.h>using namespace std;// Util function to count the elements// in first array whose atleast// one factor is present in second arrayint elementCount(int arr1[], int n1, int arr2[], int n2){ // counter to count number of elements int count = 0; // Hash of second array elements unordered_set<int> hash; for (int i = 0; i < n2; i++) hash.insert(arr2[i]); // loop to traverse through array elements // and check for its factors for (int i = 0; i < n1; i++) { // generate factors of elements // of first array for (int j = 1; j * j <= arr1[i]; j++) { // Check if j is a factor if (arr1[i] % j == 0) { // check if the factor is present in // second array using the hash if ((hash.find(j) != hash.end()) || (hash.find(arr1[i] / j) != hash.end())) { count++; break; } } } } return count;}// Driver codeint main(){ int arr1[] = { 10, 2, 13, 4, 15 }; int arr2[] = { 2, 4, 5, 6 }; int n1 = sizeof(arr1) / sizeof(arr1[0]); int n2 = sizeof(arr2) / sizeof(arr2[0]); cout << elementCount(arr1, n1, arr2, n2); return 0;} |
Java
// Java program to find count of// elements in first array whose// atleast one factor is present// in second array. import java.util.*;class GFG{ // Util function to count the elements // in first array whose atleast // one factor is present in second array static int elementCount(int arr1[], int n1, int arr2[], int n2) { // counter to count number of elements int count = 0; // Hash of second array element HashSet<Integer> hash = new HashSet<>(); for (int i = 0; i < n2; i++) { hash.add(arr2[i]); } // loop to traverse through array elements // and check for its factors for (int i = 0; i < n1; i++) { // generate factors of elements // of first array for (int j = 1; j * j <= arr1[i]; j++) { // Check if j is a factor if (arr1[i] % j == 0) { // check if the factor is present in // second array using the hash if ((hash.contains(j) && j != (int) hash.toArray()[hash.size() - 1]) || (hash.contains(arr1[i] / j) && (arr1[i] / j) != (int) hash.toArray()[hash.size() - 1])) { count++; break; } } } } return count; } // Driver code public static void main(String[] args) { int arr1[] = {10, 2, 13, 4, 15}; int arr2[] = {2, 4, 5, 6}; int n1 = arr1.length; int n2 = arr2.length; System.out.println(elementCount(arr1, n1, arr2, n2)); }}/* This code contributed by PrinciRaj1992 */ |
Python3
# Python program to find count of# elements in first array whose# atleast one factor is present# in second array. # Importing sqrt() functionfrom math import sqrt # Util function to count the # elements in first array # whose atleast one factor is# present in second arraydef elementCount(arr1, arr2): # counter to count # number of elements count = 0 # Hash of second array elements hash = frozenset(arr2) # loop to traverse through array # elements and check for its factors for x in arr1: # generate factors of # elements of first array for j in range(1, int(sqrt(x)) + 1): # Check if j is a factor if x % j == 0: # check if the factor is present # in second array using the hash if (j in hash or x / j in hash): count+=1 break return count # Driver codearr1 = [ 10, 2, 13, 4, 15 ]arr2 = [ 2, 4, 5, 6 ] print(elementCount(arr1, arr2)) # This code is contributed # by vaibhav29498 |
C#
// C# program to find count of// elements in first array whose// atleast one factor is present// in second array. using System;using System.Linq;using System.Collections.Generic;class GFG{ // Util function to count the elements // in first array whose atleast // one factor is present in second array static int elementCount(int []arr1, int n1, int []arr2, int n2) { // counter to count number of elements int count = 0; // Hash of second array element HashSet<int> hash = new HashSet<int>(); for (int i = 0; i < n2; i++) { hash.Add(arr2[i]); } // loop to traverse through array elements // and check for its factors for (int i = 0; i < n1; i++) { // generate factors of elements // of first array for (int j = 1; j * j <= arr1[i]; j++) { // Check if j is a factor if (arr1[i] % j == 0) { // check if the factor is present in // second array using the hash if ((hash.Contains(j) && j != (int) hash.ToArray()[hash.Count- 1]) || (hash.Contains(arr1[i] / j) && (arr1[i] / j) != (int) hash.ToArray()[hash.Count - 1])) { count++; break; } } } } return count; } // Driver code public static void Main(String[] args) { int []arr1 = {10, 2, 13, 4, 15}; int []arr2 = {2, 4, 5, 6}; int n1 = arr1.Length; int n2 = arr2.Length; Console.WriteLine(elementCount(arr1, n1, arr2, n2)); }}// This code contributed by Rajput-Ji |
PHP
<?php// PHP program to find count of// elements in first array whose// atleast one factor is present// in second array.// Util function to count the // elements in first array // whose atleast one factor is// present in second arrayfunction elementCount($arr1, $arr2){ // counter to count // number of elements $count = 0; // Hash of second array elements $hash = array_unique($arr2); // loop to traverse through array // elements and check for its factors foreach($arr1 as &$x) // generate factors of // elements of first array for ($j = 1; $j < (int)(sqrt($x)) + 1; $j++) // Check if j is a factor if ($x % $j == 0) { // check if the factor is present // in second array using the hash if (in_array($j, $hash) || in_array((int)($x / $j), $hash)) { $count++; break; } } return $count;}// Driver code$arr1 = array( 10, 2, 13, 4, 15 );$arr2 = array( 2, 4, 5, 6 );print(elementCount($arr1, $arr2));// This code is contributed mits?> |
Javascript
<script>// javascript program to find count of// elements in first array whose// atleast one factor is present// in second array. // Util function to count the elements // in first array whose atleast // one factor is present in second array function elementCount(arr1 , n1 , arr2 , n2) { // counter to count number of elements var count = 0; // Hash of second array element var hash = new Set(); for (i = 0; i < n2; i++) { hash.add(arr2[i]); } // loop to traverse through array elements // and check for its factors for (i = 0; i < n1; i++) { // generate factors of elements // of first array for (j = 1; j * j <= arr1[i]; j++) { // Check if j is a factor if (arr1[i] % j == 0) { // check if the factor is present in // second array using the hash if ((hash.has(j) && j != parseInt( hash[hash.length - 1]) || (hash.has(arr1[i] / j) && (arr1[i] / j) != parseInt( hash[hash.length - 1])))) { count++; break; } } } } return count; } // Driver code var arr1 = [ 10, 2, 13, 4, 15 ]; var arr2 = [ 2, 4, 5, 6 ]; var n1 = arr1.length; var n2 = arr2.length; document.write(elementCount(arr1, n1, arr2, n2));// This code contributed by umadevi9616 </script> |
Output:
4
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