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 an array which is a multiple of 4 can form an array with an 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

  Javascript

  <script>
  
  // JavaScript 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
  function findSolution(N)
  {
  
      // Length of array which is not
      // divisible by 4 is unable to
      // form such array
      if (N % 4 != 0)
          document.write(-1 + "<br>");
      else {
          let temp = 0, sum_odd = 0,
              sum_even = 0;
          let result = new Uint8Array(N);
  
          // Loop to find the resulted
          // array containing the same
          // count of even and odd elements
          for (let 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
          let diff = sum_even - sum_odd;
  
          // The difference will be added
          // in the last odd element
          result[N - 2] += diff;
  
          for (let i = 0; i < N; i++)
              document.write(result[i] + " ");
          document.write("<br>");
      }
  }
  
  // Driver Code
      let N = 8;
      findSolution(N);
  
  
  
  // This code is contributed by Surbhi Tyagi.
  </script>
• Output

  1 2 3 4 5 6 11 8
  

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