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, 11Input: 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 sequencevoid 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 Codeint main(){ int N = 12; Print(N); return 0;} |
Java
// Java Program to implement// the above approachimport 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 sequencedef 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 CodeN = 12Print(N)# This code is contributed by rohitsingh07052 |
C#
// C# Program to implement// the above approachusing 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 sequencefunction 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 Codevar 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)
