给定大小为N的数组arr [] ，任务是将数组拆分为最小数量的子集，以使每个元素恰好属于一个子集，并且可以被每个子集中存在的最小元素整除。
例子：
Input: arr[] = {10, 2, 3, 5, 4, 2}
Output: 3
Explanation:
The three possible groups are:
- {5, 10}, where all the element is divisible by 5(minimum element).
- {2, 2, 4}, where all the element is divisible by 2(minimum element).
- {3}, where all the element is divisible by 3(minimum element).
Input: arr[] = {50, 50, 50, 50, 50}
Output: 1
方法：可以通过使用“排序”并找到每个子集的最小值来解决该问题。请按照以下步骤解决问题：
- 以升序对数组arr []进行排序。
- 使用0和一个数组vis []初始化一个变量ans ，用于存储访问的数组元素。
- 将vis []数组的所有位置标记为0 ，表示尚未访问的位置。
- 遍历给定数组arr []并执行以下步骤：
- 如果未访问元素arr [i] ，则：
- 将其视为新子集的最小值，并将ans递增1 。
- 使用变量j在[i +1，N – 1]范围内进行迭代，如果元素arr [j]没有被访问且可被arr [i]整除，则将vis [j] = 1设置。
- 对每个索引重复上述步骤。
- 如果未访问元素arr [i] ，则：
- 完成上述步骤后，输出ans的值作为结果。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
#define LL long long
#define MM 1000000007
using namespace std;
// Function to find the minimum number
// of subsets into which given array
// can be split such that the given
// conditions are satisfied
void groupDivision(int arr[], int n)
{
LL z, i, j, ans;
// Sort the given array arr[]
sort(arr, arr + n);
// Initialize answer
ans = 0;
LL vis[n + 5] = { 0 };
// Iterate for the smaller value
// which has not been visited
for (i = 0; i < n; i++) {
if (!vis[i]) {
// Mark all elements that
// are divisible by arr[i]
for (j = i + 1; j < n; j++) {
// If jth index has already
// been visited
if (vis[j] == 1)
continue;
if (arr[j] % arr[i] == 0)
// Mark the jth index
// as visisted
vis[j] = 1;
}
// Increment ans by 1
ans++;
}
}
// Print the value of ans
cout << ans;
}
// Driver Code
int main()
{
int arr[] = { 10, 2, 3, 5, 4, 2 };
int N = sizeof(arr) / sizeof(arr[0]);
groupDivision(arr, N);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG
{
static int MM = 1000000007;
// Function to find the minimum number
// of subsets into which given array
// can be split such that the given
// conditions are satisfied
static void groupDivision(int arr[], int n)
{
int z, i, j, ans;
// Sort the given array arr[]
Arrays.sort(arr);
// Initialize answer
ans = 0;
int[] vis = new int[n + 5];
Arrays.fill(vis, 0);
// Iterate for the smaller value
// which has not been visited
for (i = 0; i < n; i++) {
if (vis[i] == 0) {
// Mark all elements that
// are divisible by arr[i]
for (j = i + 1; j < n; j++) {
// If jth index has already
// been visited
if (vis[j] == 1)
continue;
if (arr[j] % arr[i] == 0)
// Mark the jth index
// as visisted
vis[j] = 1;
}
// Increment ans by 1
ans++;
}
}
// Print the value of ans
System.out.println(ans);
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 10, 2, 3, 5, 4, 2 };
int N = arr.length;
groupDivision(arr, N);
}
}
// This code is contributed by code_hunt.Python3
# Python3 program for the above approach
MM = 1000000007
# Function to find the minimum number
# of subsets into which given array
# can be split such that the given
# conditions are satisfied
def groupDivision(arr, n):
global MM
ans = 0
# Sort the given array arr[]
arr = sorted(arr)
vis = [0]*(n + 5)
# Iterate for the smaller value
# which has not been visited
for i in range(n):
if (not vis[i]):
# Mark all elements that
# are divisible by arr[i]
for j in range(i + 1, n):
# If jth index has already
# been visited
if (vis[j] == 1):
continue
if (arr[j] % arr[i] == 0):
# Mark the jth index
# as visisted
vis[j] = 1
# Increment ans by 1
ans += 1
# Prthe value of ans
print (ans)
# Driver Code
if __name__ == '__main__':
arr =[10, 2, 3, 5, 4, 2]
N = len(arr)
groupDivision(arr, N)
# This code is contributed by mohit kumar 29.C#
// C# program for the above approach
using System;
class GFG
{
static int MM = 1000000007;
// Function to find the minimum number
// of subsets into which given array
// can be split such that the given
// conditions are satisfied
static void groupDivision(int[] arr, int n)
{
int z, i, j, ans;
// Sort the given array arr[]
Array.Sort(arr);
// Initialize answer
ans = 0;
int[] vis = new int[n + 5];
for (i = 0; i < n; i++) {
vis[i] = 0;
}
// Iterate for the smaller value
// which has not been visited
for (i = 0; i < n; i++) {
if (vis[i] == 0) {
// Mark all elements that
// are divisible by arr[i]
for (j = i + 1; j < n; j++) {
// If jth index has already
// been visited
if (vis[j] == 1)
continue;
if (arr[j] % arr[i] == 0)
// Mark the jth index
// as visisted
vis[j] = 1;
}
// Increment ans by 1
ans++;
}
}
// Print the value of ans
Console.Write(ans);
}
// Driver Code
static public void Main ()
{
int[] arr = { 10, 2, 3, 5, 4, 2 };
int N = arr.Length;
groupDivision(arr, N);
}
}
// This code is contributed by code_hunt.输出：
3时间复杂度： O(N 2 )
辅助空间： O(N)