给定一个整数数组arr ,任务是打印删除数组中所有元素所需的最少操作数。
在操作中,可以随机选择数组中的任何元素,并且可以从数组中删除每个可被它整除的元素。
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,应从数组中的最小元素中依次从剩余元素中选择,直到删除数组中的所有元素。
- 按升序对数组进行排序并准备出现的哈希值。
- 对于从头开始的每个未标记元素,标记所有可被选择元素整除的元素,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
PHP
Javascript
输出:
2
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