删除数组中所有元素所需的最小操作数

给定一个整数数组arr ，任务是打印删除该数组所有元素所需的最少操作数。
在操作中，可以从数组中随机选择任何元素，并且可以将其整除的每个元素从数组中删除。

例子：

Input: arr[] = {2, 4, 6, 3, 5, 10}
Output: 3
Choosing 2 as the first element will remove 2, 4, 6 and 10 from the array.
Now choose 3 which will result in the removal of 3.
Finally, the only element left to choose is 5.

Input: arr[] = {2, 5, 3, 7, 11}
Output: 5

方法：为获得最佳结果，应从其余元素中依次选择数组中最小的元素，直到删除数组中的所有元素。

• 按升序对数组进行排序，并为出现的事件准备一个哈希值。
• 对于从开始处开始的每个未标记元素，请标记所有可被select元素整除的元素，并增加结果计数器。

下面是上述方法的实现

C++

// C++ implementation of the above approach
  
#include 
#define MAX 10000
  
using namespace std;
  
int hashTable[MAX];
  
// function to find minimum operations
int minOperations(int arr[], int n)
{
    // sort array
    sort(arr, arr + n);
  
    // prepare hash of array
    for (int i = 0; i < n; i++)
        hashTable[arr[i]]++;
  
    int res = 0;
    for (int i = 0; i < n; i++) {
        if (hashTable[arr[i]]) {
            for (int j = i; j < n; j++)
                if (arr[j] % arr[i] == 0)
                    hashTable[arr[j]] = 0;
            res++;
        }
    }
  
    return res;
}
  
// Driver program
int main()
{
    int arr[] = { 4, 6, 2, 8, 7, 21, 24, 49, 44 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    cout << minOperations(arr, n);
    return 0;
}

Java

//Java implementation of the above approach 
import java.util.*;
class Solution
{
static final int MAX=10000; 
  
static int hashTable[]= new int[MAX]; 
  
// function to find minimum operations 
static int minOperations(int arr[], int n) 
{ 
    // sort array 
    Arrays.sort(arr); 
  
    // prepare hash of array 
    for (int i = 0; i < n; i++) 
        hashTable[arr[i]]++; 
  
    int res = 0; 
    for (int i = 0; i < n; i++) { 
        if (hashTable[arr[i]]!=0) { 
            for (int j = i; j < n; j++) 
                if (arr[j] % arr[i] == 0) 
                    hashTable[arr[j]] = 0; 
            res++; 
        } 
    } 
  
    return res; 
} 
  
// Driver program 
public static void main(String args[]) 
{ 
    int arr[] = { 4, 6, 2, 8, 7, 21, 24, 49, 44 }; 
    int n = arr.length; 
  
    System.out.print( minOperations(arr, n)); 
   
} 
}
// This code is contributed by Arnab Kundu

Python 3

# Python 3 implementation of 
# the above approach
MAX = 10000
  
hashTable = [0] * MAX
  
# function to find minimum operations
def minOperations(arr, n):
      
    # sort array
    arr.sort()
  
    # prepare hash of array
    for i in range(n):
        hashTable[arr[i]] += 1
  
    res = 0
    for i in range(n) :
        if (hashTable[arr[i]]) :
            for j in range(i, n):
                if (arr[j] % arr[i] == 0):
                    hashTable[arr[j]] = 0
            res += 1
  
    return res
  
# Driver Code
if __name__ == "__main__":
    arr = [ 4, 6, 2, 8, 7, 21, 24, 49, 44 ]
    n = len(arr)
  
    print(minOperations(arr, n))
  
# This code is contributed 
# by ChitraNayal

C#

using System;
  
// C# implementation of the above approach  
public class Solution
{
public const int MAX = 10000;
  
public static int[] hashTable = new int[MAX];
  
// function to find minimum operations  
public static int minOperations(int[] arr, int n)
{
    // sort array  
    Array.Sort(arr);
  
    // prepare hash of array  
    for (int i = 0; i < n; i++)
    {
        hashTable[arr[i]]++;
    }
  
    int res = 0;
    for (int i = 0; i < n; i++)
    {
        if (hashTable[arr[i]] != 0)
        {
            for (int j = i; j < n; j++)
            {
                if (arr[j] % arr[i] == 0)
                {
                    hashTable[arr[j]] = 0;
                }
            }
            res++;
        }
    }
  
    return res;
}
  
// Driver program  
public static void Main(string[] args)
{
    int[] arr = new int[] {4, 6, 2, 8, 7, 21, 24, 49, 44};
    int n = arr.Length;
  
    Console.Write(minOperations(arr, n));
  
}
}
  
// This code is contributed by Shrikant13

输出：

2