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Print N distinct numbers following the given operations
• Last Updated : 20 Apr, 2021

Given an integer N which is always an even number, our task is to print N distinct numbers following the given conditions:

• The first half numbers are even whereas the other half numbers are odd
• Sum of the elements of first half numbers and sum of elements of second half numbers should be equal

Print the array if the above conditions satisfy otherwise output “-1”.
Examples:

Input: N = 4
Output: 2 4 1 5
Explanation:
Given number 4 we are required to print 4 numbers. First Half = 2, 4 and their sum is 6, other half = 1, 5 and their sum is also 6.
Input: N = 22
Output: -1
Explanation:
It is not possible to print the required array.

Approach:
To solve the problem mentioned above we have to observe that the integer N has to be a multiple of 4

• We know that the sum of the first N/2 even numbers will be even, so if the sum of the other N/2 integers is also even then N/2 must be even, because the sum of an odd number of odd integers is always odd.
• If N/2 is even then N is a multiple of 4, so if n is not divisible by 4 then the answer is “-1”, otherwise, there will be a possible array.
• For printing the array we will consider two parts such that the first half that is N/2 elements will be simply multiples of 2 and the other half will be multiple of 2 – 1. For the last element in the array, we will calculate the integer by applying the direct formula N + N / 2 – 1 because we are supposed to make the sum for two halves equal.

Below is the implementation of the above approach:

## C++

 `// C++ implementation to Print N distinct numbers``#include ``using` `namespace` `std;` `// Function to print the required array``bool` `printArr(``int` `n)``{` `    ``// Check if number is a multiple of 4``    ``if` `(n % 4 == 0) {``        ``// Printing Left Half of the array``        ``for` `(``int` `i = 1; i <= n / 2; i++)``            ``cout << i * 2 << ``' '``;` `        ``// Printing Right Half of the array``        ``for` `(``int` `i = 1; i < n / 2; i++)``            ``cout << i * 2 - 1 << ``' '``;``        ``cout << n + n / 2 - 1 << ``'\n'``;``    ``}` `    ``else``        ``cout << ``"-1"``;``}` `// Driver code``int` `main()``{``    ``int` `n = 22;` `    ``printArr(n);` `    ``return` `0;``}`

## Java

 `// Java implementation to print N distinct numbers``import` `java.util.*;` `class` `GFG{` `// Function to print the required array``static` `void` `printArr(``int` `n)``{` `    ``// Check if number is a multiple of 4``    ``if``(n % ``4` `== ``0``)``    ``{``       ``// Printing left half of the array``       ``for``(``int` `i = ``1``; i <= n / ``2``; i++)``           ``System.out.print(i * ``2` `+ ``" "``);` `       ``// Printing Right Half of the array``       ``for``(``int` `i = ``1``; i < n / ``2``; i++)``           ``System.out.print(i * ``2` `- ``1` `+ ``" "``);` `       ``System.out.println(n + n / ``2` `- ``1``);``    ``}``    ``else``        ``System.out.print(``"-1"``);``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `n = ``22``;``    ``printArr(n);``}``}` `// This code is contributed by amal kumar choubey`

## Python3

 `# Python3 implementation to print``# N distinct numbers` `# Function to print the required array``def` `printArr(n):` `    ``# Check if number is a multiple of 4``    ``if` `(n ``%` `4` `=``=` `0``):``        ` `        ``# Printing Left Half of the array``        ``for` `i ``in` `range``(``1``, (n ``/` `2``) ``+` `1``):``            ``print` `(i ``*` `2``, end ``=` `" "``)` `        ``# Printing Right Half of the array``        ``for` `i ``in` `range``(``1``, n ``/` `2``):``            ``print` `(i ``*` `2` `-` `1``, end ``=` `" "``)``            ` `        ``print` `(n ``+` `n ``/` `2` `-` `1``, end ``=` `"\n"``)` `    ``else``:``        ``print` `(``"-1"``)` `# Driver code``n ``=` `22``printArr(n)` `# This code is contributed by PratikBasu`

## C#

 `// C# implementation to print N distinct numbers``using` `System;` `public` `class` `GFG{` `// Function to print the required array``static` `void` `printArr(``int` `n)``{` `    ``// Check if number is a multiple of 4``    ``if``(n % 4 == 0)``    ``{``        ` `    ``// Printing left half of the array``    ``for``(``int` `i = 1; i <= n / 2; i++)``        ``Console.Write(i * 2 + ``" "``);` `    ``// Printing Right Half of the array``    ``for``(``int` `i = 1; i < n / 2; i++)``        ``Console.Write(i * 2 - 1 + ``" "``);` `    ``Console.WriteLine(n + n / 2 - 1);``    ``}``    ` `    ``else``        ``Console.Write(``"-1"``);``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``int` `n = 22;``    ``printArr(n);``}``}` `// This code is contributed by 29AjayKumar`

## Javascript

 ``
Output:
`-1`

Time Complexity: O(N)

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