Construct an Array of size N in which sum of odd elements is equal to sum of even elements

Given an integer N which is always even, the task is to create an array of size N which contains N/2 even numbers and N/2 odd numbers. All the elements of array should be distinct and the sum of even numbers is equal to the sum of odd numbers. If no such array exists then print -1.
Examples: 
 

Input: N = 8 
Output: 2 4 6 8 1 3 5 11 
Explanation: 
The array has 8 distinct elements which have equal sum of odd and even numbers, i.e., (2 + 4 + 6 + 8 = 1 + 3 + 5 + 11).
Input: N = 10 
Output: -1 
Explanation: 
It is not possible to construct array of size 10. 
 

Approach: To solve the problem mentioned above the very first observation is that it is not possible to create an array that has size N which is a multiple of 2 but not multiple of 4. Because, if that happens then the sum of one half which contains odd numbers will always be odd and the sum of another half which contains even numbers will always be even, hence the sum of both halves can't be the same.
Therefore, the array which satisfies the problem statement should always have a size N which is a multiple of 4. The approach is to first construct N/2 even numbers starting from 2, which is the first half of the array. Then create another part of the array starting from 1 and finally calculate the last odd element such that it makes both the halves equal. In order to do so the last element of odd numbers should be (N/2) - 1 + N.
Below is the implementation of the above approach: 
 

// C++ program to Create an array
// of size N consisting of distinct
// elements where sum of odd elements
// is equal to sum of even elements

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

// Function to construct the required array
void arrayConstruct(int N)
{

    // To construct first half,
    // distinct even numbers
    for (int i = 2; i <= N; i = i + 2)
        cout << i << " ";

    // To construct second half,
    // distinct odd numbers
    for (int i = 1; i < N - 1; i = i + 2)
        cout << i << " ";

    // Calculate the last number of second half
    // so as to make both the halves equal
    cout << N - 1 + (N / 2) << endl;
}

// Function to construct the required array
void createArray(int N)
{

    // check if size is multiple of 4
    // then array exist
    if (N % 4 == 0)

        // function call to construct array
        arrayConstruct(N);

    else
        cout << -1 << endl;
}

// Driver code
int main()
{

    int N = 8;

    createArray(N);

    return 0;
}

// Java program to Create an array
// of size N consisting of distinct
// elements where sum of odd elements
// is equal to sum of even elements
class GFG{
 
// Function to construct the required array
static void arrayConstruct(int N)
{
 
    // To confirst half,
    // distinct even numbers
    for (int i = 2; i <= N; i = i + 2)
        System.out.print(i+ " ");
 
    // To consecond half,
    // distinct odd numbers
    for (int i = 1; i < N - 1; i = i + 2)
        System.out.print(i+ " ");
 
    // Calculate the last number of second half
    // so as to make both the halves equal
    System.out.print(N - 1 + (N / 2) +"\n");
}
 
// Function to construct the required array
static void createArray(int N)
{
 
    // check if size is multiple of 4
    // then array exist
    if (N % 4 == 0)
 
        // function call to conarray
        arrayConstruct(N);
 
    else
        System.out.print(-1 +"\n");
}
 
// Driver code
public static void main(String[] args)
{
    int N = 8;
 
    createArray(N);
}
}

// This code is contributed by Princi Singh

# python3 program to Create an array
# of size N consisting of distinct
# elements where sum of odd elements
# is equal to sum of even elements

# Function to construct the required array
def arrayConstruct(N):

    # To construct first half,
    # distinct even numbers
    for i in range(2, N + 1, 2):
        print(i,end=" ")

    # To construct second half,
    # distinct odd numbers
    for i in range(1, N - 1, 2):
        print(i, end=" ")

    # Calculate the last number of second half
    # so as to make both the halves equal
    print(N - 1 + (N // 2))

# Function to construct the required array
def createArray(N):

    # check if size is multiple of 4
    # then array exist
    if (N % 4 == 0):

        # function call to construct array
        arrayConstruct(N)

    else:
        cout << -1 << endl

# Driver code
if __name__ == '__main__':

    N = 8

    createArray(N)

# This code is contributed by mohit kumar 29

// C# program to Create an array
// of size N consisting of distinct
// elements where sum of odd elements
// is equal to sum of even elements
using System;

class GFG{
  
// Function to construct the required array
static void arrayConstruct(int N)
{
  
    // To confirst half,
    // distinct even numbers
    for (int i = 2; i <= N; i = i + 2)
        Console.Write(i + " ");
  
    // To consecond half,
    // distinct odd numbers
    for (int i = 1; i < N - 1; i = i + 2)
        Console.Write(i + " ");
  
    // Calculate the last number of second half
    // so as to make both the halves equal
    Console.Write(N - 1 + (N / 2) +"\n");
}
  
// Function to construct the required array
static void createArray(int N)
{
  
    // check if size is multiple of 4
    // then array exist
    if (N % 4 == 0)
  
        // function call to conarray
        arrayConstruct(N);
  
    else
        Console.Write(-1 +"\n");
}
  
// Driver code
public static void Main(String[] args)
{
    int N = 8;
  
    createArray(N);
}
}

// This code is contributed by Rajput-Ji

<script>
// JavaScript program to Create an array
// of size N consisting of distinct
// elements where sum of odd elements
// is equal to sum of even elements

// Function to construct the required array
function arrayConstruct(N)
{

    // To construct first half,
    // distinct even numbers
    for (let i = 2; i <= N; i = i + 2)
        document.write(i + " ");

    // To construct second half,
    // distinct odd numbers
    for (let i = 1; i < N - 1; i = i + 2)
        document.write(i + " ");

    // Calculate the last number of second half
    // so as to make both the halves equal
    document.write(N - 1 + (N / 2) + "<br>");
}

// Function to construct the required array
function createArray(N)
{

    // check if size is multiple of 4
    // then array exist
    if (N % 4 == 0)

        // function call to construct array
        arrayConstruct(N);
    else
        document.write(-1 + "<br>");
}

// Driver code
    let N = 8;
    createArray(N);

// This code is contributed by Surbhi Tyagi.
</script>

2 4 6 8 1 3 5 11

Time Complexity: O(N)

Auxiliary Space: O(1)