Minimum elements inserted in a sorted array to form an Arithmetic progression

Given a sorted array arr[], the task is to find minimum elements needed to be inserted in the array such that array forms an Arithmetic Progression.
Examples:

Input: arr[] = {1, 6, 8, 10, 14, 16} 
Output: 10 
Explanation:
Minimum elements required to form A.P. is 10. 
Transformed array after insertion of the elements. 
arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 
9, 10, 11, 12, 13, 14, 15, 16} 
Input: arr[] = {1, 3, 5, 7, 11} 
Output: 1 
Explanation:
Minimum elements required to form A.P. is 1. 
Transformed array after insertion of the elements. 
arr[] = {1, 3, 5, 7, 9, 11} 

Approach: The idea is to find the difference of consecutive elements of the sorted array and then find the greatest common divisor of all the differences. The GCD of the differences will be the common-difference for the arithmetic progression that can be formed. Below is the illustration of the steps:

• Find the difference between consecutive elements of the array and store in it diff[].
• Now the GCD of the diff[] array will give the common difference between the elements of given sorted array. 
  For Example:
  

Given array be {1, 5, 7}
Difference of Consecutive elements will be -
Difference(1, 5) = |5 - 1| = 4
Difference(5, 7) = |7 - 5| = 2

Then, GCD of the Differences will be 
gcd(4, 2) = 2

This means there can be A.P. formed
with common-difference as 2. That is - 
{1, 3, 5, 7}

• If the difference between consecutive elements of the sorted array arr[] is greater than the GCD calculated above, then the minimum number of elements needed to be inserted in the given array to make the element of the array in Arithmetic Progression is given by: 
  

Number of possible elements = 
    (Difference / Common Difference) - 1

• Add the count of element needed to be insert for all set of consecutive elements in the given sorted array.

Below is the implementation of the above approach:

C++

// C++ implementation to find the
// minimum elements required to
// be inserted into an array to
// form an arithmetic progression

#include <bits/stdc++.h>
using namespace std;

// Function to find the greatest
// common divisor of two numbers
int gcdFunc(int a, int b)
{
    if (b == 0)
        return a;
    return gcdFunc(b, a % b);
}

// Function to find the minimum
// the minimum number of elements
// required to be inserted into array
int findMinimumElements(int* a, int n)
{
    int b[n - 1];

    // Difference array of consecutive
    // elements of the array
    for (int i = 1; i < n; i++) {
        b[i - 1] = a[i] - a[i - 1];
    }
    int gcd = b[0];

    // GCD of the difference array
    for (int i = 0; i < n - 1; i++) {
        gcd = gcdFunc(gcd, b[i]);
    }
    int ans = 0;

    // Loop to calculate the minimum
    // number of elements required
    for (int i = 0; i < n - 1; i++) {
        ans += (b[i] / gcd) - 1;
    }
    return ans;
}

// Driver Code
int main()
{
    int arr1[] = { 1, 6, 8, 10, 14, 16 };
    int n1 = sizeof(arr1)/sizeof(arr1[0]);
    // Function calling
    cout << findMinimumElements(arr1, n1)
         << endl;
}

Java

// Java implementation to find the
// minimum elements required to
// be inserted into an array to
// form an arithmetic progression

class GFG{

    // Function to find the greatest
    // common divisor of two numbers
    static int gcdFunc(int a, int b)
    {
        if (b == 0)
            return a;
        return gcdFunc(b, a % b);
    }

    // Function to find the minimum
    // the minimum number of elements
    // required to be inserted into array
    static int findMinimumElements(int[] a, int n)
    {
        int[] b = new int[n - 1];

        // Difference array of consecutive
        // elements of the array
        for (int i = 1; i < n; i++) {
            b[i - 1] = a[i] - a[i - 1];
        }
        int gcd = b[0];

        // GCD of the difference array
        for (int i = 0; i < n - 1; i++) {
            gcd = gcdFunc(gcd, b[i]);
        }
        int ans = 0;

        // Loop to calculate the minimum
        // number of elements required
        for (int i = 0; i < n - 1; i++) {
            ans += (b[i] / gcd) - 1;
        }
        return ans;
    }

 // Driver Code
public static void main(String[] args)
{
    int arr1[] = { 1, 6, 8, 10, 14, 16 };
    int n1 = arr1.length;
    // Function calling
    System.out.print(findMinimumElements(arr1, n1)
         +"\n");
}
}

// This code is contributed by Rajput-Ji

Python3

# Python3 implementation to find the
# minimum elements required to
# be inserted into an array to
# form an arithmetic progression

# Function to find the greatest
# common divisor of two numbers
def gcdFunc(a, b):
    if (b == 0):
        return a

    return gcdFunc(b, a % b)

# Function to find the minimum
# the minimum number of elements
# required to be inserted into array
def findMinimumElements(a, n):
    b = [0]*(n - 1)

    # Difference array of consecutive
    # elements of the array
    for i in range(1,n):
        b[i - 1] = a[i] - a[i - 1]

    gcd = b[0]

    # GCD of the difference array
    for i in range(n-1):
        gcd = gcdFunc(gcd, b[i])

    ans = 0

    # Loop to calculate the minimum
    # number of elements required
    for i in range(n-1):
        ans += (b[i] // gcd) - 1

    return ans

# Driver Code
arr1 = [1, 6, 8, 10, 14, 16]
n1 = len(arr1)
# Function calling
print(findMinimumElements(arr1, n1))

# This code is contributed by shubhamsingh10

C#

// C# implementation to find the
// minimum elements required to
// be inserted into an array to
// form an arithmetic progression
using System;

class GFG{

    // Function to find the greatest
    // common divisor of two numbers
    static int gcdFunc(int a, int b)
    {
        if (b == 0)
            return a;
        return gcdFunc(b, a % b);
    }

    // Function to find the minimum
    // the minimum number of elements
    // required to be inserted into array
    static int findMinimumElements(int[] a, int n)
    {
        int[] b = new int[n - 1];

        // Difference array of consecutive
        // elements of the array
        for (int i = 1; i < n; i++) {
            b[i - 1] = a[i] - a[i - 1];
        }
        int gcd = b[0];

        // GCD of the difference array
        for (int i = 0; i < n - 1; i++) {
            gcd = gcdFunc(gcd, b[i]);
        }
        int ans = 0;

        // Loop to calculate the minimum
        // number of elements required
        for (int i = 0; i < n - 1; i++) {
            ans += (b[i] / gcd) - 1;
        }
        return ans;
    }

    // Driver Code
    static public void Main ()
    {
        int[] arr1 = new int[] { 1, 6, 8, 10, 14, 16 };
        int n1 = arr1.Length;
        // Function calling
        Console.WriteLine(findMinimumElements(arr1, n1));
    }
}

// This code is contributed by shivanisingh

Javascript

<script>

// Javascript implementation to find the
// minimum elements required to
// be inserted into an array to
// form an arithmetic progression


// Function to find the greatest
// common divisor of two numbers
function gcdFunc(a, b)
{
    if (b == 0)
        return a;
    return gcdFunc(b, a % b);
}

// Function to find the minimum
// the minimum number of elements
// required to be inserted into array
function findMinimumElements(a, n)
{
    let b = new Array(n - 1);

    // Difference array of consecutive
    // elements of the array
    for (let i = 1; i < n; i++) {
        b[i - 1] = a[i] - a[i - 1];
    }
    let gcd = b[0];

    // GCD of the difference array
    for (let i = 0; i < n - 1; i++) {
        gcd = gcdFunc(gcd, b[i]);
    }
    let ans = 0;

    // Loop to calculate the minimum
    // number of elements required
    for (let i = 0; i < n - 1; i++) {
        ans += (b[i] / gcd) - 1;
    }
    return ans;
}

// Driver Code

    let arr1 = [ 1, 6, 8, 10, 14, 16 ];
    let n1 = arr1.length;
    // Function calling
    document.write(findMinimumElements(arr1, n1)
        + "<br>");

// This code is contributed by Mayank Tyagi

</script>

Output:

10

Time Complexity: O(N * log(MAX)), where N is the number of elements in the array and MAX is the maximum element in the array.
Auxiliary Space: O(N + log(MAX))