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# 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)

Article Tags :

Analysis of Algorithms

C++

Competitive Programming

DSA

Mathematical

Programming Language

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