# Count elements that are divisible by at-least one element in another array

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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `using` `namespace` `std; ` ` `  `// Util function to count the elements ` `// in first array whose atleast ` `// one factor is present in second array ` `int` `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 code ` `int` `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 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() function ` `from` `math ``import` `sqrt ` `  `  `# Util function to count the  ` `# elements in first array  ` `# whose atleast one factor is ` `# present in second array ` `def` `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 code ` `arr1 ``=` `[ ``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

 ` `

Output:

```4
```

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