Maximum difference elements that can added to a set

Given a set containing N elements, you are allowed to add an element Z > 0 to this set only if it can be represented as |X-Y| where X and Y are already present in the set. After adding an element Z, you can use it as an element of the set to add new elements. Find out the maximum number of elements that can be added in this way.

Examples:

Input : set = {2, 3}
Output : 1
The only element that can be added is 1.

Input : set = {4, 6, 10}
Output : 2
The 2 elements that can be added are 
(6-4) = 2 and (10-2) = 8.

This problem is based on the following observations:



Below is the implementation of the above approach:

filter_none

edit
close

play_arrow

link
brightness_4
code

// CPP program to find the maximum number
// of elements that can be added to a set
// such that it is the absolute difference
// of 2 elements already in the set
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the maximum number
// of elements that can be added to a set
// such that it is the absolute difference
// of 2 elements already in the set
int maxNewElements(int a[], int n)
{
    int gcd = a[0];
  
    int mx = a[0];
  
    for (int i = 1; i < n; i++) {
        gcd = __gcd(gcd, a[i]);
        mx = max(mx, a[i]);
    }
  
    int total_terms = mx / gcd;
  
    return total_terms - n;
}
  
// Driver Code
int main()
{
    int a[] = { 4, 6, 10 };
    int n = sizeof(a) / sizeof(a[0]);
    cout << maxNewElements(a, n);
    return 0;
}
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to find the maximum number
// of elements that can be added to a set
// such that it is the absolute difference
// of 2 elements already in the set
   
import java.util.*;
import java.lang.*;
import java.io.*;
  
  
class GFG{
      
static int __gcd(int a, int b) {
   if (b==0) return a;
   return __gcd(b,a%b);
}
// Function to find the maximum number
// of elements that can be added to a set
// such that it is the absolute difference
// of 2 elements already in the set
static int maxNewElements(int []a, int n)
{
    int gcd = a[0];
   
    int mx = a[0];
   
    for (int i = 1; i < n; i++) {
        gcd = __gcd(gcd, a[i]);
        mx = Math.max(mx, a[i]);
    }
   
    int total_terms = mx / gcd;
   
    return total_terms - n;
}
   
// Driver Code
public static void main(String args[])
{
    int a[] = { 4, 6, 10 };
    int n = a.length;
    System.out.print(maxNewElements(a, n));
}
}
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 program to find the maximum number 
# of elements that can be added to a set 
# such that it is the absolute difference 
# of 2 elements already in the set 
  
# from math lib import gcd method
from math import gcd
  
  
# Function to find the maximum number 
# of elements that can be added to a set 
# such that it is the absolute difference 
# of 2 elements already in the set 
def maxNewElements(a, n) :
  
    __gcd = a[0]
  
    mx = a[0]
  
    for i in range(1, n) :
        __gcd = gcd(__gcd,a[i])
        mx = max(mx, a[i])
  
    total_terms = mx / __gcd
  
    return total_terms - n
  
  
  
  
# Driver code
if __name__ == "__main__" :
  
    a = [ 4, 6, 10 ]
  
    n = len(a)
  
    print(int(maxNewElements(a,n)))
  
# This code is contributed by 
# ANKITRAI1
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to find the maximum 
// number of elements that can be 
// added to a set such that it is 
// the absolute difference of 2 
// elements already in the set
class GFG
{
      
static int __gcd(int a, int b) 
{
    if (b == 0) return a;
    return __gcd(b, a % b);
}
  
// Function to find the maximum number
// of elements that can be added to a set
// such that it is the absolute difference
// of 2 elements already in the set
static int maxNewElements(int[] a, int n)
{
    int gcd = a[0];
  
    int mx = a[0];
  
    for (int i = 1; i < n; i++)
    {
        gcd = __gcd(gcd, a[i]);
        mx = System.Math.Max(mx, a[i]);
    }
  
    int total_terms = mx / gcd;
  
    return total_terms - n;
}
  
// Driver Code
static void Main()
{
    int[] a = { 4, 6, 10 };
    int n = a.Length;
    System.Console.WriteLine(maxNewElements(a, n));
}
}
  
// This code is contributed by mits
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP program to find the maximum number
// of elements that can be added to a set
// such that it is the absolute difference
// of 2 elements already in the set
  
function __gcd($a, $b
{
    if ($b == 0) 
        return $a;
    return __gcd($b, $a % $b);
}
  
// Function to find the maximum number
// of elements that can be added to 
// a set such that it is the absolute 
// difference of 2 elements already in
// the set
function maxNewElements($a, $n)
{
    $gcd = $a[0];
  
    $mx = $a[0];
  
    for ($i = 1; $i < $n; $i++)
    {
        $gcd = __gcd($gcd, $a[$i]);
        $mx = max($mx, $a[$i]);
    }
  
    $total_terms = $mx / $gcd;
  
    return $total_terms - $n;
}
  
// Driver Code
$a = array(4, 6, 10 );
$n = sizeof($a);
echo maxNewElements($a, $n);
  
// This code is contributed 
// by Akanksha Rai
chevron_right

Output:
2

Time Complexity: O(N*LogN)




Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.





Article Tags :
Practice Tags :