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.
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.
C++
#include <iostream>
using namespace std;
void findNumbers( int n, int d)
{
for ( int i = 0; i < n - 2; i++)
cout << "1" << " " ;
cout << "2" << " " ;
cout << n + d << endl;
}
int main()
{
int N = 3, D = 5;
findNumbers(N, D);
return 0;
}
|
Java
import java.io.*;
class GFG {
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);
}
public static void main(String args[])
{
int N = 3 , D = 5 ;
findNumbers(N, D);
}
}
|
Python3
def pattern(n, d) :
for i in range ( 0 , n - 2 ) :
print ( "1" , end = " " )
print ( "2" , end = " " )
print (n + d)
N = 3
D = 5
pattern(N, D)
|
C#
using System;
class GFG {
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);
}
public static void Main()
{
int N = 3, D = 5;
findNumbers(N, D);
}
}
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PHP
<?php
function findNumbers( $n , $d )
{
for ( $i = 0; $i < $n - 2; $i ++)
echo "1" , " " ;
echo "2" , " " ;
echo $n + $d , "\n" ;
}
$N = 3;
$D = 5;
findNumbers( $N , $D );
?>
|
Javascript
<script>
function findNumbers(n, d)
{
for (let i = 0; i < n - 2; i++)
document.write( "1" + " " );
document.write( "2" + " " );
document.write(n + d);
}
let N = 3, D = 5;
findNumbers(N, D);
</script>
|
Time complexity : O(n)
Auxiliary Space : O(1)
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