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

• Last Updated : 22 Apr, 2021

Given an integer N (3 â‰¤ N â‰¤ 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:
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 od elements will always be odd. Therefore, it is not possible to generate such an array.

Approach: Follow the steps below to solve the problem:

1. If N is odd or (N / 2) is odd, then print -1.
2. 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 ``using` `namespace` `std;` `// Fun dtion 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``{` `  ``// Fun dtion 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` `# Fun dtion 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#

 `// C# Program to implement``// the above approach``using` `System;``class` `GFG``{` `  ``// Fun dtion 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

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

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

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