# Maximize count of distinct elements possible in an Array from the given operation

Given an array A[] of size N, the task is to maximize the count of distinct elements in the array by inserting the absolute differences of the existing array elements.
Examples:

Input: A[] = { 1, 2, 3, 5 }
Output:
Explanation:
Possible absolute differences among the array elements are:
(2 – 1) = 1
(5 – 3) = 2
(5 – 2) = 3
(5 – 1) = 4
Hence, inserting 4 into the array maximizes the count of distinct elements.
Input: A[] = { 1, 2, 3, 6 }
Output:

Naive Approach:
Generate all possible pairs from the array and store their absolute differences in a set. Insert all the array elements into the set. The final size of the set denotes the maximum distinct elements that the array can possess after performing given operations.
Time Complexity: O(N2
Auxiliary Space: O(N)
Efficient Approach:
Follow the steps below to optimize the above approach:

1. Find the maximum element of the array.The maximum possible absolute difference of any two elements does not exceed the maximum element of the array.

2. Calculate GCD of the array, since it is the common factor of the whole array.

3. Divide the maximum element with the gcd of the array to get the count of the distinct elements.

Below is the implementation of the above approach:

## C++

 `// C++ program to find the maximum ` `// possible distinct elements that ` `// can be obtained from the array ` `// by performing the given operations ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find the gcd ` `// of two numbers ` `int` `gcd(``int` `x, ``int` `y) ` `{ ` `    ``if` `(x == 0) ` `        ``return` `y; ` `    ``return` `gcd(y % x, x); ` `} ` ` `  `// Function to calculate and return ` `// the count of maximum possible ` `// distinct elements in the array ` `int` `findDistinct(``int` `arr[], ``int` `n) ` `{ ` `    ``// Find the maximum element ` `    ``int` `maximum = *max_element(arr, ` `                               ``arr + n); ` ` `  `    ``// Base Case ` `    ``if` `(n == 1) ` `        ``return` `1; ` ` `  `    ``if` `(n == 2) { ` `        ``return` `(maximum / gcd(arr, ` `                              ``arr)); ` `    ``} ` ` `  `    ``// Finding the gcd of first ` `    ``// two element ` `    ``int` `k = gcd(arr, arr); ` ` `  `    ``// Calculate Gcd of the array ` `    ``for` `(``int` `i = 2; i < n; i++) { ` `        ``k = gcd(k, arr[i]); ` `    ``} ` ` `  `    ``// Return the total count ` `    ``// of distinct elements ` `    ``return` `(maximum / k); ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 2, 3, 5 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``cout << findDistinct(arr, n); ` `    ``return` `0; ` `} `

## Java

 `// Java program to find the maximum ` `// possible distinct elements that ` `// can be obtained from the array ` `// by performing the given operations ` `import` `java.util.*; ` ` `  `class` `GFG{ ` `     `  `// Function to find the gcd ` `// of two numbers ` `static` `int` `gcd(``int` `x, ``int` `y) ` `{ ` `    ``if` `(x == ``0``) ` `        ``return` `y; ` `    ``return` `gcd(y % x, x); ` `} ` ` `  `// Function to calculate and return ` `// the count of maximum possible ` `// distinct elements in the array ` `static` `int` `findDistinct(``int` `arr[], ``int` `n) ` `{ ` `     `  `    ``// Find the maximum element ` `    ``int` `maximum = Arrays.stream(arr).max().getAsInt();  ` ` `  `    ``// Base Case ` `    ``if` `(n == ``1``) ` `        ``return` `1``; ` ` `  `    ``if` `(n == ``2``) ` `    ``{ ` `        ``return` `(maximum / gcd(arr[``0``], ` `                              ``arr[``1``])); ` `    ``} ` ` `  `    ``// Finding the gcd of first ` `    ``// two element ` `    ``int` `k = gcd(arr[``0``], arr[``1``]); ` ` `  `    ``// Calculate Gcd of the array ` `    ``for``(``int` `i = ``2``; i < n; i++)  ` `    ``{ ` `        ``k = gcd(k, arr[i]); ` `    ``} ` ` `  `    ``// Return the total count ` `    ``// of distinct elements ` `    ``return` `(maximum / k); ` `} ` ` `  `// Driver code  ` `public` `static` `void` `main (String[] args)  ` `{ ` `    ``int` `arr[] = { ``1``, ``2``, ``3``, ``5` `}; ` `    ``int` `n = arr.length; ` ` `  `    ``System.out.println(findDistinct(arr, n)); ` `} ` `} ` ` `  `// This code is contributed by offbeat `

## Python3

 `# Python3 program to find the maximum ` `# possible distinct elements that ` `# can be obtained from the array ` `# by performing the given operations ` ` `  `# Function to find the gcd ` `# of two numbers ` `def` `gcd(x, y): ` `     `  `    ``if` `(x ``=``=` `0``): ` `        ``return` `y ` `    ``return` `gcd(y ``%` `x, x) ` ` `  `# Function to calculate and return ` `# the count of maximum possible ` `# distinct elements in the array ` `def` `findDistinct(arr, n): ` `     `  `    ``# Find the maximum element ` `    ``maximum ``=` `max``(arr) ` `     `  `    ``# Base Case ` `    ``if` `(n ``=``=` `1``): ` `        ``return` `1` ` `  `    ``if` `(n ``=``=` `2``): ` `        ``return` `(maximum ``/``/` `gcd(arr[``0``],  ` `                               ``arr[``1``])) ` ` `  `    ``# Finding the gcd of first ` `    ``# two element ` `    ``k ``=` `gcd(arr[``0``], arr[``1``]) ` ` `  `    ``# Calculate Gcd of the array ` `    ``for` `i ``in` `range``(``2``, n): ` `        ``k ``=` `gcd(k, arr[i]) ` ` `  `    ``# Return the total count ` `    ``# of distinct elements ` `    ``return` `(maximum ``/``/` `k) ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `'__main__'``: ` ` `  `    ``arr ``=` `[ ``1``, ``2``, ``3``, ``5` `] ` `    ``n ``=` `len``(arr) ` ` `  `    ``print``(findDistinct(arr, n)) ` ` `  `# This code is contributed by mohit kumar 29 `

## C#

 `// C# program to find the maximum ` `// possible distinct elements that ` `// can be obtained from the array ` `// by performing the given operations ` `using` `System; ` `using` `System.Linq; ` `class` `GFG{ ` ` `  `// Function to find the gcd ` `// of two numbers ` `static` `int` `gcd(``int` `x, ``int` `y) ` `{ ` `    ``if` `(x == 0) ` `        ``return` `y; ` `    ``return` `gcd(y % x, x); ` `} ` ` `  `// Function to calculate and return ` `// the count of maximum possible ` `// distinct elements in the array ` `static` `int` `findDistinct(``int` `[]arr, ``int` `n) ` `{ ` `    ``// Find the maximum element ` `    ``int` `maximum = arr.Max(); ` ` `  `    ``// Base Case ` `    ``if` `(n == 1) ` `        ``return` `1; ` ` `  `    ``if` `(n == 2) ` `    ``{ ` `        ``return` `(maximum / gcd(arr, ` `                              ``arr)); ` `    ``} ` ` `  `    ``// Finding the gcd of first ` `    ``// two element ` `    ``int` `k = gcd(arr, arr); ` ` `  `    ``// Calculate Gcd of the array ` `    ``for` `(``int` `i = 2; i < n; i++)  ` `    ``{ ` `        ``k = gcd(k, arr[i]); ` `    ``} ` ` `  `    ``// Return the total count ` `    ``// of distinct elements ` `    ``return` `(maximum / k); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main() ` `{ ` `    ``int` `[]arr = { 1, 2, 3, 5 }; ` `    ``int` `n = arr.Length; ` ` `  `    ``Console.Write(findDistinct(arr, n)); ` `} ` `} ` ` `  `// This code is contributed by Code_Mech `

Output:

```5
```

Time Complexity: O(N)
Auxiliary Space: O(1)

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