将数组拆分为最小数量的子集，其中子集的每个元素都可以被其最小值整除

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。
• 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.

Javascript

 
// Javascript program for the above approach
var MM = 1000000007;
 
// Creating the bblSort function
function bblSort(arr)
{
    for(var i = 0; i < arr.length; i++)
    {
     
        // Last i elements are already in place 
        for(var j = 0;
                j < (arr.length - i - 1);
                j++)
        {
         
            // Checking if the item at present iteration
            // is greater than the next iteration
            if (arr[j] > arr[j + 1])
            {
             
                // If the condition is true
                // then swap them
                var temp = arr[j]
                arr[j] = arr[j + 1]
                arr[j + 1] = temp
            }
        }
    }
     
    // Return the sorted array
    return (arr);
}
 
// Function to find the minimum number
// of subsets into which given array
// can be split such that the given
// conditions are satisfied
function groupDivision(arr, n)
{
    var z, i, j, ans;
 
    // Sort the given array arr
    arr = bblSort(arr);
 
    // Initialize answer
    ans = 0;
    var vis = Array(n + 5).fill(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
    document.write(ans);
}
 
// Driver Code
var arr = [ 10, 2, 3, 5, 4, 2 ];
var N = arr.length;
 
groupDivision(arr, N);
 
// This code is contributed by gauravrajput1
 

输出：

3

时间复杂度： O(N ² )
辅助空间： O(N)