Generate an N-length array having equal count and sum of elements of both parities

Last Updated : 28 Nov, 2022

Given an integer N (3 ≤ N ≤ 10⁵), the task is to generate an array of N distinct positive elements such that the count and sum of elements of both parties i.e even and odd, are the same. If it is not possible to construct such an array, print -1.

Examples:

Input: N = 8
Output: 2, 4, 6, 8, 1, 3, 5, 11

Input: 6
Output: -1
Explanation: For the count of odd and even array elements to be equal, 3 even and odd elements must be present in the array. But, sum of 3 odd elements will always be odd. Therefore, it is not possible to generate such an array.

Approach: Follow the steps below to solve the problem:

If N is odd or (N / 2) is odd, then print -1.
Otherwise:
- Print N/2 even values starting from 2 and store the count in some variable, say SumEven.
- Print N / 2 - 1 odd values starting from 1 and store the sum in another variable, say SumOdd.
- At last print SumEven - SumOdd which is also odd.

Below is the implementation of the above approach:

C++

// C++ Program to implement // the above approach  #include <bits/stdc++.h> using namespace std;  // Function to print the // required sequence void Print(int N) {     if ((N / 2) % 2 == 1  || (N % 2 == 1)) {         cout << -1 << endl;         return;     }  // Stores count of even // and odd elements     int CurEven = 2, CurOdd = 1;  // Stores sum of even // and odd elements     int SumOdd = 0, SumEven = 0;  // Print N / 2 even elements     for (int i = 0; i < (N / 2); i++) {          cout << CurEven << " ";         SumEven += CurEven;         CurEven += 2;     }  // Print N / 2 - 1 odd elements     for (int i = 0; i < N / 2 - 1; i++) {         cout << CurOdd << " ";         SumOdd += CurOdd;         CurOdd += 2;     }      CurOdd = SumEven - SumOdd;      // Print final odd element     cout << CurOdd << '\n'; }  // Driver Code int main() {     int N = 12;     Print(N);     return 0; }

Java

// Java Program to implement // the above approach import java.io.*; class GFG  {    // Function to print the   // required sequence   static void Print(int N)   {     if ((N / 2) % 2 == 1 || (N % 2 == 1))      {       System.out.print(-1);       return;     }      // Stores count of even     // and odd elements     int CurEven = 2, CurOdd = 1;      // Stores sum of even     // and odd elements     int SumOdd = 0, SumEven = 0;      // Print N / 2 even elements     for (int i = 0; i < (N / 2); i++)      {       System.out.print(CurEven + " ");       SumEven += CurEven;       CurEven += 2;     }      // Print N / 2 - 1 odd elements     for (int i = 0; i < N / 2 - 1; i++)     {       System.out.print(CurOdd + " ");       SumOdd += CurOdd;       CurOdd += 2;     }     CurOdd = SumEven - SumOdd;      // Print final odd element     System.out.println(CurOdd);   }    // Driver Code   public static void main (String[] args)    {     int N = 12;     Print(N);   } }  // This code is contributed by Dharanendra L V.

Python3

# Python Program to implement # the above approach  # Function to print the # required sequence def Print(N):     if ((N / 2) % 2 or (N % 2)):         print(-1)         return  # Stores count of even # and odd elements     CurEven = 2     CurOdd = 1  # Stores sum of even # and odd elements     SumOdd = 0     SumEven = 0  # Print N / 2 even elements     for i in range(N // 2):          print(CurEven,end=" ")         SumEven += CurEven         CurEven += 2  # Print N / 2 - 1 odd elements     for i in range( (N//2) - 1):         print(CurOdd, end=" ")         SumOdd += CurOdd         CurOdd += 2      CurOdd = SumEven - SumOdd      # Print final odd element     print(CurOdd)  # Driver Code N = 12 Print(N)  # This code is contributed by rohitsingh07052

// C# Program to implement // the above approach using System; class GFG  {    // Function to print the   // required sequence   static void Print(int N)   {     if ((N / 2) % 2 == 1 || (N % 2 == 1))      {       Console.WriteLine(-1);       return;     }      // Stores count of even     // and odd elements     int CurEven = 2, CurOdd = 1;      // Stores sum of even     // and odd elements     int SumOdd = 0, SumEven = 0;      // Print N / 2 even elements     for (int i = 0; i < (N / 2); i++)      {       Console.Write(CurEven + " ");       SumEven += CurEven;       CurEven += 2;     }      // Print N / 2 - 1 odd elements     for (int i = 0; i < N / 2 - 1; i++)     {       Console.Write(CurOdd + " ");       SumOdd += CurOdd;       CurOdd += 2;     }     CurOdd = SumEven - SumOdd;      // Print final odd element     Console.WriteLine(CurOdd);   }    // Driver Code   public static void Main()   {     int N = 12;     Print(N);   } }  // This code is contributed by chitranayal.

JavaScript

<script>  // Javascript Program to implement // the above approach  // Function to print the // required sequence function Print(N) {     if ((N / 2) % 2 == 1 || (N % 2 == 1))      {          document.write(-1);         return;     }          // Stores count of even     // and odd elements     var CurEven = 2, CurOdd = 1;          // Stores sum of even     // and odd elements     var SumOdd = 0, SumEven = 0;          // Print N / 2 even elements     for(var i = 0; i < (N / 2); i++)      {         document.write(CurEven + " ");         SumEven += CurEven;         CurEven += 2;     }          // Print N / 2 - 1 odd elements     for(var i = 0; i < N / 2 - 1; i++)     {         document.write(CurOdd + " ");         SumOdd += CurOdd;         CurOdd += 2;     }     CurOdd = SumEven - SumOdd;          // Print final odd element     document.write(CurOdd); }  // Driver Code var N = 12;  Print(N);  // This code is contributed by Ankita saini  </script>

Output:

2 4 6 8 10 12 1 3 5 7 9 17

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

Count subarrays having sum of elements at even and odd positions equal

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Generate an N-length array having equal count and sum of elements of both parities

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