满足给定条件的数组可能的最大子集

给定一个数组arr[]和一个整数K 。任务是找到最大子集的大小，使得子集(X, Y)中的每一对都具有Y != (X * K) 形式，其中X < Y 。

例子：

Input: arr[] = {2, 3, 6, 5, 4, 10}, K = 2 
Output: 3 
{2, 3, 5} is the required sub-set

Input: arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, K = 2 
Output: 6 

方法：

• 对所有数组元素进行排序。
• 创建一个空的整数集S ，它将保存子集的元素。
• 遍历已排序的数组，对于数组中的每个整数x ：
  • 如果x % k = 0或x / k不存在于S 中，则将x插入到S 中。
  • 否则丢弃x并检查下一个元素。
• 最后打印集合S的大小。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the size of the required sub-set
int sizeSubSet(int a[], int k, int n)
{
    // Sort the array
    sort(a, a + n);
 
    // Set to store the contents of the required sub-set
    unordered_set s;
 
    // Insert the elements satisfying the conditions
    for (int i = 0; i < n; i++) {
        if (a[i] % k != 0 || s.count(a[i] / k) == 0)
            s.insert(a[i]);
    }
 
    // Return the size of the set
    return s.size();
}
 
// Driver code
int main()
{
    int a[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
    int n = sizeof(a) / sizeof(a[0]);
    int k = 2;
 
    cout << sizeSubSet(a, k, n);
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG
{
     
// Function to return the size of the required sub-set
static int sizeSubSet(int a[], int k, int n)
{
    // Sort the array
    Arrays.sort(a);
 
    // HashMap to store the contents
    // of the required sub-set
    HashMap< Integer, Integer> s = new HashMap< Integer, Integer>();
     
    // Insert the elements satisfying the conditions
    for (int i = 0; i < n; i++)
    {
        if (a[i] % k != 0 || s.get(a[i] / k) == null)
            s.put(a[i], s.get(a[i]) == null ? 1 : s.get(a[i]) + 1);
    }
 
    // Return the size of the set
    return s.size();
}
 
// Driver code
public static void main(String args[])
{
    int a[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
    int n = a.length;
    int k = 2;
    System.out.println( sizeSubSet(a, k, n));
}
}
// This code is contributed by Arnab Kundu

Python3

# Python3 implementation of the approach
 
import math as mt
# Function to return the size of the required sub-set
def sizeSubSet(a, k, n):
 
    # Sort the array
    a.sort()
  
    # Set to store the contents of the required sub-set
    s=set()
  
    # Insert the elements satisfying the conditions
    for i in range(n):
        if (a[i] % k != 0 or a[i] // k not in s):
            s.add(a[i])
     
  
    # Return the size of the set
    return len(s)
 
  
# Driver code
a=[1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ]
n = len(a)
k = 2
 
print(sizeSubSet(a, k, n))
 
# This is contributed by Mohit kumar 29

C#

// C# mplementation of the approach
using System;
using System.Collections.Generic;
 
class GFG
{
     
// Function to return the size of
// the required sub-set
static int sizeSubSet(int []a, int k, int n)
{
    // Sort the array
    Array.Sort(a);
 
    // HashMap to store the contents
    // of the required sub-set
    Dictionary s = new Dictionary();
     
    // Insert the elements satisfying the conditions
    for (int i = 0; i < n; i++)
    {
        if (a[i] % k != 0 || !s.ContainsKey(a[i] / k))
        {
            if(s.ContainsKey(a[i]))
            {
                var val = s[a[i]];
                s.Remove(a[i]);
                s.Add(a[i], val + 1);
            }
            else
            {
                s.Add(a[i], 1);
            }
        }
    }
 
    // Return the size of the set
    return s.Count;
}
 
// Driver code
public static void Main(String []args)
{
    int []a = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
    int n = a.Length;
    int k = 2;
    Console.WriteLine(sizeSubSet(a, k, n));
}
}
 
// This code is contributed by PrinciRaj1992

Javascript

 
// Javascript implementation of the approach
 
// Function to return the size of the
// required sub-set
function sizeSubSet(a, k, n)
{
     
    // Sort the array
    a.sort(function(a, b){return a - b;});
   
    // HashMap to store the contents
    // of the required sub-set
    let s = new Map();
       
    // Insert the elements satisfying the conditions
    for(let i = 0; i < n; i++)
    {
        if (a[i] % k != 0 ||
            s.get(a[i] / k) == null)
            s.set(a[i], s.get(a[i]) == null ?
                    1 : s.get(a[i]) + 1);
    }
   
    // Return the size of the set
    return s.size;
}
 
// Driver code
let a = [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ];
let n = a.length;
let k = 2;
 
document.write(sizeSubSet(a, k, n));
 
// This code is contributed by patel2127
 

输出：

6

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