Largest sub-set possible for an array satisfying the given condition

Given an array arr[] and an integer K. The task is to find the size of the maximum sub-set such that every pair from the sub-set (X, Y) is of the form Y != (X * K) where X < Y.

Examples:

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 

Approach:

• Sort all the array elements.
• Create an empty set of integers S, which will hold the elements for the sub-set.
• Traverse the sorted array, and for each integer x in the array: 
  • If x % k = 0 or x / k is not already present in S then insert x into S.
  • Else discard x and check the next element.
• Print the size of the set S in the end.

Below is the implementation of the above approach:  

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
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<int> 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# implementation 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<int,
               int> s = new Dictionary<int,
                                       int>();

    // 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

PHP

<?php
// Php implementation of the approach

// Function to return the size of
// the required sub-set
function sizeSubSet($a, $k, $n)
{

    // Sort the array
    sort($a) ;

    // Set to store the contents of
    // the required sub-set
    $s = array();

    // Insert the elements satisfying
    // the conditions
    for ($i = 0 ; $i < $n ; $i++)
    {
        if ($a[$i] % $k != 0 or
            !in_array(floor($a[$i] / $k), $s))
            array_push($s, $a[$i]);
    }

    // Return the size of the set
    return sizeof($s);

}

// Driver code
$a = array(1, 2, 3, 4, 5, 6, 7, 8, 9, 10 );
$n = sizeof($a);
$k = 2;

echo sizeSubSet($a, $k, $n);

// This code is contributed by Ryuga
?>

Javascript

<script>

// 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

</script>

Output:

6

Time Complexity: O(n*log(n)), As we are sorting the array
Auxiliary Space: O(n)