Find the Array which when sorted forms an AP and has minimum maximum

Last Updated : 29 Nov, 2021

Given three positive integers N, X, and Y(X<Y). The task is to find an array of length N containing both X and Y, and when sorted in increasing order, the array must form an arithmetic progression

Examples:

Input: N = 5, X = 20, Y = 50
Output: 20 30 40 50 10
Explanation: The array when sorted in increasing order forms an arithmetic progression with common difference 10.

Input: N = 17, X = 23445, Y = 1000000
Output: 23445 218756 414067 609378 804689 1000000 1195311 1390622 1585933 1781244 1976555 2171866 2367177 2562488 2757799 2953110 3148421
Explanation: The array when sorted in increasing order forms an arithmetic progression with common difference 195311.

Approach: In this problem, it can be observed that if the maximum element is to be minimized, then it can be assumed that Y should be the greatest element. If Y is the greatest, then each of the remaining elements will be less than or equal to Y. If Y is not the greatest element in the array, then elements greater than Y can be considered.

Follow the steps below to solve the given problem:

Check if some elements which have a common difference are present between X and Y. For this, check if (Y-X) is divisible by (N-1). If it is divisible, then decrease N and the common difference d is given by:

d = (Y-X)/(N-1)

If it is not divisible, then decrease (n-1) and repeat the above step in a loop till the denominator is non-zero.
If there don't exist any such elements between X and Y, then the common difference d is given by:

d = (Y-X)

Let the resultant array be stored in vector ans. Push the elements found between X and Y in the vector ans to use a for loop.
If N is not equal to zero, insert the elements in ans starting from (X-d) with common difference d.
If N is equal to zero, output ans. Otherwise, insert the elements in ans starting from (Y+d) till the required array is obtained.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

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

// Function to find the array which when
// sorted forms an arithmetic progression
// and has the minimum possible maximum element
void findArray(int N, int X, int Y)
{
    // Stores the difference of Y and X
    int r = Y - X;

    // To indicate the absence of required
    // elements between X and Y
    int d = -1;

    // Stores the number of required elements
    // present between X and Y
    int num = 0;

    // Loop to find the number of required
    // elements between X and Y
    for (int i = (N - 1); i > 0; i--) {

        // If r is divisible by i
        if (r % i == 0) {

            // Stores the common difference
            // of the elements
            d = r / i;

            // Decrease num by 1
            num = i - 1;

            // Break from the loop
            break;
        }
    }

    // If the number of elements present
    // between X and Y is zero
    if (d == -1) {

        // Update common difference
        d = Y - X;
    }

    // Update N
    N = N - 2 - num;

    // Stores the resultant array
    vector<int> ans;

    // Loop to insert the found elements
    // in the resultant vector
    for (int i = X; i <= Y; i += d) {
        ans.push_back(i);
    }

    // Stores the element to be inserted in
    // the resultant vector
    int i = X;

    // Loop to insert elements less than X
    // in the resultant vector
    while (N > 0 && i > 0) {
        i = i - d;

        // i must be positive
        if (i > 0) {
            ans.push_back(i);

            // Update N
            N--;
        }
    }

    // Stores the element to be inserted in
    // the resultant vector
    i = Y;

    // Loop to insert elements greater than Y
    // in the resultant vector
    while (N > 0) {
        i = i + d;
        ans.push_back(i);

        // Update N
        N--;
    }

    // Output the required array
    for (int i = 0; i < ans.size(); ++i) {
        cout << ans[i] << " ";
    }
}

// Driver Code
int main()
{
    // Given N, X and Y
    int N = 5, X = 20, Y = 50;

    // Function Call
    findArray(N, X, Y);

    return 0;
}

Java

// Java program for the above approach
import java.util.ArrayList;

class GFG {

    // Function to find the array which when
    // sorted forms an arithmetic progression
    // and has the minimum possible maximum element
    static void findArray(int N, int X, int Y) {

        // Stores the difference of Y and X
        int r = Y - X;

        // To indicate the absence of required
        // elements between X and Y
        int d = -1;

        // Stores the number of required elements
        // present between X and Y
        int num = 0;

        // Loop to find the number of required
        // elements between X and Y
        for (int i = (N - 1); i > 0; i--) {

            // If r is divisible by i
            if (r % i == 0) {

                // Stores the common difference
                // of the elements
                d = r / i;

                // Decrease num by 1
                num = i - 1;

                // Break from the loop
                break;
            }
        }

        // If the number of elements present
        // between X and Y is zero
        if (d == -1) {

            // Update common difference
            d = Y - X;
        }

        // Update N
        N = N - 2 - num;

        // Stores the resultant array
        ArrayList<Integer> ans = new ArrayList<Integer>();

        // Loop to insert the found elements
        // in the resultant vector
        for (int i = X; i <= Y; i += d) {
            ans.add(i);
        }

        // Stores the element to be inserted in
        // the resultant vector
        int j = X;

        // Loop to insert elements less than X
        // in the resultant vector
        while (N > 0 && j > 0) {
            j = j - d;

            // i must be positive
            if (j > 0) {
                ans.add(j);

                // Update N
                N--;
            }
        }

        // Stores the element to be inserted in
        // the resultant vector
        j = Y;

        // Loop to insert elements greater than Y
        // in the resultant vector
        while (N > 0) {
            j = j + d;
            ans.add(j);

            // Update N
            N--;
        }

        // Output the required array
        for (int i = 0; i < ans.size(); ++i) {
            System.out.print(ans.get(i) + " ");
        }
    }

    // Driver Code
    public static void main(String[] args) 
    {
      
        // Given N, X and Y
        int N = 5, X = 20, Y = 50;

        // Function Call
        findArray(N, X, Y);
    }
}

 // This code is contributed by gfgking.

Python3

# python program for the above approach

# Function to find the array which when
# sorted forms an arithmetic progression
# and has the minimum possible maximum element
def findArray(N, X, Y):

    # Stores the difference of Y and X
    r = Y - X

    # To indicate the absence of required
    # elements between X and Y
    d = -1

    # Stores the number of required elements
    # present between X and Y
    num = 0

    # Loop to find the number of required
    # elements between X and Y
    for i in range(N - 1, 0, -1):

        # If r is divisible by i
        if (r % i == 0):

            # Stores the common difference
            # of the elements
            d = r // i

            # Decrease num by 1
            num = i - 1

            # Break from the loop
            break

    # If the number of elements present
    # between X and Y is zero
    if (d == -1):

        # Update common difference
        d = Y - X

    # Update N
    N = N - 2 - num

    # Stores the resultant array
    ans = []

    # Loop to insert the found elements
    # in the resultant vector
    for i in range(X, Y + 1, d):
        ans.append(i)

    # Stores the element to be inserted in
    # the resultant vector
    i = X

    # Loop to insert elements less than X
    # in the resultant vector
    while (N > 0 and i > 0):
        i = i - d

        # i must be positive
        if (i > 0):
            ans.append(i)

            # Update N
            N -= 1

    # Stores the element to be inserted in
    # the resultant vector
    i = Y

    # Loop to insert elements greater than Y
    # in the resultant vector
    while (N > 0):
        i = i + d
        ans.append(i)

        # Update N
        N -= 1

    # Output the required array
    for i in range(0, len(ans)):
        print(ans[i], end=" ")

# Driver Code
if __name__ == "__main__":

    # Given N, X and Y
    N = 5
    X = 20
    Y = 50

    # Function Call
    findArray(N, X, Y)

    # This code is contributed by rakeshsahni

// C# program for the above approach
using System;
using System.Collections.Generic;

public class GFG
{
  
// Function to find the array which when
// sorted forms an arithmetic progression
// and has the minimum possible maximum element
static void findArray(int N, int X, int Y)
{
  
    // Stores the difference of Y and X
    int r = Y - X;

    // To indicate the absence of required
    // elements between X and Y
    int d = -1;

    // Stores the number of required elements
    // present between X and Y
    int num = 0;

    // Loop to find the number of required
    // elements between X and Y
    for (int i = (N - 1); i > 0; i--) {

        // If r is divisible by i
        if (r % i == 0) {

            // Stores the common difference
            // of the elements
            d = r / i;

            // Decrease num by 1
            num = i - 1;

            // Break from the loop
            break;
        }
    }

    // If the number of elements present
    // between X and Y is zero
    if (d == -1) {

        // Update common difference
        d = Y - X;
    }

    // Update N
    N = N - 2 - num;

    // Stores the resultant array
    List<int> ans = new List<int>();

    // Loop to insert the found elements
    // in the resultant vector
    for (int i = X; i <= Y; i += d) {
        ans.Add(i);
    }

    // Stores the element to be inserted in
    // the resultant vector
    int j = X;

    // Loop to insert elements less than X
    // in the resultant vector
    while (N > 0 && j > 0) {
        j = j - d;

        // i must be positive
        if (j > 0) {
            ans.Add(j);

            // Update N
            N--;
        }
    }

    // Stores the element to be inserted in
    // the resultant vector
    j = Y;

    // Loop to insert elements greater than Y
    // in the resultant vector
    while (N > 0) {
        j = j + d;
        ans.Add(j);

        // Update N
        N--;
    }

    // Output the required array
    for (int i = 0; i < ans.Count; ++i) {
        Console.Write( ans[i] + " ");
    }
}

// Driver Code
public static void Main(String[] args)
{
    // Given N, X and Y
    int N = 5, X = 20, Y = 50;

    // Function Call
    findArray(N, X, Y);
}
}

// This code is contributed by code_hunt.

JavaScript

 <script>

        // JavaScript Program to implement
        // the above approach 

        // Function to find the array which when
        // sorted forms an arithmetic progression
        // and has the minimum possible maximum element
        function findArray(N, X, Y)
        {
        
            // Stores the difference of Y and X
            let r = Y - X;

            // To indicate the absence of required
            // elements between X and Y
            let d = -1;

            // Stores the number of required elements
            // present between X and Y
            let num = 0;

            // Loop to find the number of required
            // elements between X and Y
            for (let i = (N - 1); i > 0; i--) {

                // If r is divisible by i
                if (r % i == 0) {

                    // Stores the common difference
                    // of the elements
                    d = r / i;

                    // Decrease num by 1
                    num = i - 1;

                    // Break from the loop
                    break;
                }
            }

            // If the number of elements present
            // between X and Y is zero
            if (d == -1) {

                // Update common difference
                d = Y - X;
            }

            // Update N
            N = N - 2 - num;

            // Stores the resultant array
            let ans = [];

            // Loop to insert the found elements
            // in the resultant vector
            for (let i = X; i <= Y; i += d) {
                ans.push(i);
            }

            // Stores the element to be inserted in
            // the resultant vector
            let i = X;

            // Loop to insert elements less than X
            // in the resultant vector
            while (N > 0 && i > 0) {
                i = i - d;

                // i must be positive
                if (i > 0) {
                    ans.push(i);

                    // Update N
                    N--;
                }
            }

            // Stores the element to be inserted in
            // the resultant vector
            i = Y;

            // Loop to insert elements greater than Y
            // in the resultant vector
            while (N > 0) {
                i = i + d;
                ans.push(i);

                // Update N
                N--;
            }

            // Output the required array
            for (let i = 0; i < ans.length; ++i) {
                document.write(ans[i] + " ");
            }
        }

        // Driver Code

        // Given N, X and Y
        let N = 5, X = 20, Y = 50;

        // Function Call
        findArray(N, X, Y);

    // This code is contributed by Potta Lokesh
    </script>

Output

20 30 40 50 10

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

Find the Array which when sorted forms an AP and has minimum maximum

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Find the Array which when sorted forms an AP and has minimum maximum

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