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:
- The maximum element that can be inserted has to be less than the current maximum element in the set, since the difference of 2 integers can not exceed any of the integers.
- Every number that you insert has to be a multiple of the gcd of the given array. Since at any step, X and Y both are multiples of gcd, (X-Y) will also be a multiple of the gcd.
- You can insert all multiples of gcd less than the maximum element.
- The number of terms less than or equal to max divisible by gcd is floor (max/gcd), which will be the total number of elements in the set after performing all insertions, we need to remove the count of original N elements to get our answer.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
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;
}
int main()
{
int a[] = { 4, 6, 10 };
int n = sizeof (a) / sizeof (a[0]);
cout << maxNewElements(a, n);
return 0;
}
|
Java
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);
}
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;
}
public static void main(String args[])
{
int a[] = { 4 , 6 , 10 };
int n = a.length;
System.out.print(maxNewElements(a, n));
}
}
|
Python 3
from math import gcd
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
if __name__ = = "__main__" :
a = [ 4 , 6 , 10 ]
n = len (a)
print ( int (maxNewElements(a,n)))
|
C#
class GFG
{
static int __gcd( int a, int b)
{
if (b == 0) return a;
return __gcd(b, a % b);
}
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;
}
static void Main()
{
int [] a = { 4, 6, 10 };
int n = a.Length;
System.Console.WriteLine(maxNewElements(a, n));
}
}
|
PHP
<?php
function __gcd( $a , $b )
{
if ( $b == 0)
return $a ;
return __gcd( $b , $a % $b );
}
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 ;
}
$a = array (4, 6, 10 );
$n = sizeof( $a );
echo maxNewElements( $a , $n );
|
Javascript
<script>
function __gcd(a, b)
{
if (b == 0) return a;
return __gcd(b, a % b);
}
function maxNewElements(a, n)
{
let gcd = a[0];
let mx = a[0];
for (let i = 1; i < n; i++)
{
gcd = __gcd(gcd, a[i]);
mx = Math.max(mx, a[i]);
}
let total_terms = parseInt(mx / gcd, 10);
return total_terms - n;
}
let a = [ 4, 6, 10 ];
let n = a.length;
document.write(maxNewElements(a, n));
</script>
|
Time Complexity: O(N*LogN)
Auxiliary Space: O(1)
Last Updated :
25 Aug, 2022
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