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 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<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() 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
<?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 array function 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