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: 5 
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: 6  

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 the 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++ program to find the maximum
// possible distinct elements that
// can be obtained from the array
// by performing the given operations
#include <bits/stdc++.h>
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[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
int main()
{
    int arr[] = { 1, 2, 3, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);

    cout << findDistinct(arr, n);
    return 0;
}

// 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 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# 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[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()
{
    int []arr = { 1, 2, 3, 5 };
    int n = arr.Length;

    Console.Write(findDistinct(arr, n));
}
}

// This code is contributed by Code_Mech

<script>

// Javascript 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
function 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
function findDistinct(arr, n)
{
      
    // Find the maximum element
    let maximum = Math.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
    let k = gcd(arr[0], arr[1]);
  
    // Calculate Gcd of the array
    for(let i = 2; i < n; i++) 
    {
        k = gcd(k, arr[i]);
    }
  
    // Return the total count
    // of distinct elements
    return (maximum / k);
}

// Driver Code
    
    let arr = [ 1, 2, 3, 5 ];
    let n = arr.length;
  
    document.write(findDistinct(arr, n));

</script>

5

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