给定两个数组 arr1[] 和 arr2[]。任务是在第一个数组中找到至少一个因子出现在第二个数组中的元素的计数。
例子：
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
这个想法是将第二个数组的所有元素插入到一个散列中，以便可以在恒定时间内完成对因子的查找。现在，遍历第一个数组，为每个元素生成从 1 开始的所有因子，并检查哈希中是否存在任何因子。
下面是上述方法的实现：
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 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 vaibhav29498C#
// 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 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.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
输出：
4
如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。