Wednesday, July 3, 2024
HomeData ModellingData Structure & AlgorithmLargest sub-set possible for an array satisfying the given condition

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:
{2, 3, 5} is the required sub-set

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

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)

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!

RELATED ARTICLES

LEAVE A REPLY

Please enter your comment!
Please enter your name here

Most Popular

Recent Comments