Generate an array of given size with equal count and sum of odd and even numbers

Given an integer N, the task is to find an array of length N that contains same count of odd and even elements with an equal sum of even and odd elements in the array.

Note: Print -1 if no such array is possible.
Examples:

Input: N = 4
Output: 1 2 5 4
Explanation:
Even elements of the array – {2, 4}, S(even) = 6
Odd elements of the array – {1, 5}, S(odd) = 6

Input: N = 6
Output: -1
Explanation:
There are no such array which contains 3 even elements and 3 odd elements with equal sum.

Approach: The key observation in the problem is that only the length of array which is multiple of 4 can form array with equal number of even and odd elements with equal sum. Below is the illustration of the steps:

• Even elements of the array is the first N/2 even elements of the natural numbers starting from 2.
• Similarly, (N/2 – 1) odd elements of the array is the first (N/2 – 1) odd elements of the natural numbers starting from 1.
• The Last odd element of the array is the required value to make the sum of the even and odd elements of the array equal.

  Last Odd Element = 
     (sum of even elements) - 
     (sum of N/2 - 1 odd elements)
  

Below is the implementation of the above approach:

C++

// C++ implementation to find the
// array containing same count of
// even and odd elements with equal
// sum of even and odd elements

#include <bits/stdc++.h>

using namespace std;

// Function to find the array such that
// the array contains the same count
// of even and odd elements with equal
// sum of even and odd elements
void findSolution(int N)
{

    // Length of array which is not
    // divisible by 4 is unable to
    // form such array
    if (N % 4 != 0)
        cout << -1 << "\n";
    else {
        int temp = 0, sum_odd = 0,
            sum_even = 0;
        int result[N] = { 0 };

        // Loop to find the resulted
        // array containing the same
        // count of even and odd elements
        for (int i = 0; i < N; i += 2) {
            temp += 2;

            result[i + 1] = temp;
            // Find the total sum
            // of even elements
            sum_even += result[i + 1];

            result[i] = temp - 1;
            // Find the total sum
            // of odd elements
            sum_odd += result[i];
        }

        // Find the difference between the
        // total sum of even and odd elements
        int diff = sum_even - sum_odd;

        // The difference will be added
        // in the last odd element
        result[N - 2] += diff;

        for (int i = 0; i < N; i++)
            cout << result[i] << " ";
        cout << "\n";
    }
}

// Driver Code
int main()
{
    int N = 8;
    findSolution(N);
    return 0;
}

Java

// Java implementation to find the
// array containing same count of
// even and odd elements with equal
// sum of even and odd elements

class GFG{

// Function to find the array such that
// the array contains the same count
// of even and odd elements with equal
// sum of even and odd elements
static void findSolution(int N)
{

    // Length of array which is not
    // divisible by 4 is unable to
    // form such array
    if (N % 4 != 0)
        System.out.print(-1 + "\n");

    else
    {
        int temp = 0, sum_odd = 0;
        int sum_even = 0;
        int result[] = new int[N];

        // Loop to find the resulted
        // array containing the same
        // count of even and odd elements
        for(int i = 0; i < N; i += 2)
        {
           temp += 2;
           result[i + 1] = temp;

           // Find the total sum
           // of even elements
           sum_even += result[i + 1];
           result[i] = temp - 1;

           // Find the total sum
           // of odd elements
           sum_odd += result[i];
        }

        // Find the difference between the
        // total sum of even and odd elements
        int diff = sum_even - sum_odd;

        // The difference will be added
        // in the last odd element
        result[N - 2] += diff;

        for(int i = 0; i < N; i++)
           System.out.print(result[i] + " ");
        System.out.print("\n");
    }
}

// Driver Code
public static void main(String[] args)
{
    int N = 8;
    findSolution(N);
}
}

// This code is contributed by Amit Katiyar

Python3

# Python3 implementation to find the
# array containing same count of
# even and odd elements with equal
# sum of even and odd elements

# Function to find the array such that
# the array contains the same count
# of even and odd elements with equal
# sum of even and odd elements
def findSolution(N):

    # Length of array which is not
    # divisible by 4 is unable to
    # form such array
    if (N % 4 != 0):
        print(-1)
    else:
        temp = 0
        sum_odd = 0
        sum_even = 0
        result = [0] * N

        # Loop to find the resulted
        # array containing the same
        # count of even and odd elements
        for i in range(0, N, 2):
            temp += 2
            result[i + 1] = temp

            # Find the total sum
            # of even elements
            sum_even += result[i + 1]
            result[i] = temp - 1

            # Find the total sum
            # of odd elements
            sum_odd += result[i]

        # Find the difference between the
        # total sum of even and odd elements
        diff = sum_even - sum_odd

        # The difference will be added
        # in the last odd element
        result[N - 2] += diff

        for i in range(N):
            print(result[i], end = " ")
        print()

# Driver Code
N = 8;
findSolution(N)

# This code is contributed by divyamohan123

C#

// C# implementation to find the
// array containing same count of
// even and odd elements with equal
// sum of even and odd elements
using System;

class GFG{

// Function to find the array such that
// the array contains the same count
// of even and odd elements with equal
// sum of even and odd elements
static void findSolution(int N)
{

    // Length of array which is not
    // divisible by 4 is unable to
    // form such array
    if (N % 4 != 0)
        Console.Write(-1 + "\n");

    else
    {
        int temp = 0, sum_odd = 0;
        int sum_even = 0;
        int []result = new int[N];

        // Loop to find the resulted
        // array containing the same
        // count of even and odd elements
        for(int i = 0; i < N; i += 2)
        {
           temp += 2;
           result[i + 1] = temp;

           // Find the total sum
           // of even elements
           sum_even += result[i + 1];
           result[i] = temp - 1;

           // Find the total sum
           // of odd elements
           sum_odd += result[i];
        }

        // Find the difference between the
        // total sum of even and odd elements
        int diff = sum_even - sum_odd;

        // The difference will be added
        // in the last odd element
        result[N - 2] += diff;

        for(int i = 0; i < N; i++)
           Console.Write(result[i] + " ");
        Console.Write("\n");
    }
}

// Driver Code
public static void Main(String[] args)
{
    int N = 8;
    findSolution(N);
}
}

// This code is contributed by Rohit_ranjan

Output:

1 2 3 4 5 6 11 8

Performance Analysis:

• Time Complexity: O(N)
• Auxiliary Space: O(1)