Find N integers with given difference between product and sum

Given two integers N and D, Find a set of N integers such that the difference between their product and sum is equal to D.

Examples:  

Input : N = 2, D = 1
Output : 2 3
Explanation: 
product = 2*3 = 6,
Sum = 2 + 3 = 5.
Hence, 6 - 5 = 1(D).

Input : N = 3, D = 5.
Output : 1 2 8
Explanation :
Product = 1*2*8 = 16
Sum = 1+2+8 = 11.
Hence, 16-11 = 5(D).

A tricky solution is to keep the difference D to choose N numbers as N-2 '1's, one '2' and one remaining number as 'N+D'. 
Sum = (N-2)*(1) + 2 + (N+D) = 2*N + D. 
Product = 1*2*(N+D) = 2*N+2*D 
Difference = (2*N+2*D) - (2*N+D) = D. 

// CPP code to generate numbers
// with difference between
// product and sum is D
#include <iostream>
using namespace std;

// Function to implement calculation
void findNumbers(int n, int d)
{
    for (int i = 0; i < n - 2; i++)
        cout << "1"  << " ";

    cout << "2" << " ";
    cout << n + d << endl;
}

// Driver code
int main()
{
    int N = 3, D = 5;
    findNumbers(N, D);
    return 0;
}

// Java code to generate numbers
// with difference between
// product and sum is D
import java.io.*;

class GFG {
    
    // Function to implement calculation
    static void findNumbers(int n, int d)
    {
        for (int i = 0; i < n - 2; i++)
            System.out.print("1" + " ");
    
        System.out.print("2" + " ");
        System.out.println(n + d);
    }
    
    // Driver code
    public static void main(String args[])
    {
        int N = 3, D = 5;
        findNumbers(N, D);
    }
}

/* This code is contributed by Nikita Tiwari.*/

# Python3 code to generate numbers with
# difference between product and sum is D

# Function to implement calculation
def pattern(n, d) :
    
    for i in range(0, n - 2) :
        print("1", end=" ")
        
    print("2", end=" ")
    print(n + d)

# Driver code
N = 3
D = 5
pattern(N, D)


# This code is contributed by 'Akanshgupta'

// C# code to generate numbers
// with difference between
// product and sum is D
using System;

class GFG {
    
    // Function to implement calculation
    static void findNumbers(int n, int d)
    {
        for (int i = 0; i < n - 2; i++)
        Console.Write("1" + " ");
    
        Console.Write("2" + " ");
        Console.Write(n + d);
    }
    
    // Driver code
    public static void Main()
    {
        int N = 3, D = 5;
        findNumbers(N, D);
    }
}

/* This code is contributed by vt_m.*/

<?php
// PHP code to generate numbers
// with difference between
// product and sum is D

// Function to implement
// calculation
function findNumbers($n, $d)
{
    for ($i = 0; $i < $n - 2; $i++)
        echo "1" ," ";

    echo "2" , " ";
    echo $n + $d ,"\n";
}

    // Driver Code
    $N = 3; 
    $D = 5;
    findNumbers($N, $D);

// This code is contributed by ajit
?>

<script>

// JavaScript program to generate numbers
// with difference between
// product and sum is D

// Function to implement calculation
    function findNumbers(n, d)
    {
        for (let i = 0; i < n - 2; i++)
            document.write("1" + " ");
      
        document.write("2" + " ");
       document.write(n + d);
    }
      

// Driver code

        let N = 3, D = 5;
        findNumbers(N, D);
           
</script>

Output : 

 1 2 8

Time complexity : O(n) 
Auxiliary Space : O(1)