# 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.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// 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 ` `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; ` ` `  `    ``int` `mx = a; ` ` `  `    ``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); ` `    ``cout << maxNewElements(a, n); ` `    ``return` `0; ` `} `

## Java

 `// 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)); ` `} ` `} `

## Python 3

 `# 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 `

## C#

 `// 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; ` ` `  `    ``int` `mx = a; ` ` `  `    ``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 `

## PHP

 `

Output:

```2
```

Time Complexity: O(N*LogN)

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