Given an integer N which is always even, the task is to create an array of size N which contains N/2 even numbers and N/2 odd numbers. All the elements of array should be distinct and the sum of even numbers is equal to the sum of odd numbers. If no such array exists then print -1.
Examples:
Input: N = 8
Output: 2 4 6 8 1 3 5 11
Explanation:
The array has 8 distinct elements which have equal sum of odd and even numbers, i.e., (2 + 4 + 6 + 8 = 1 + 3 + 5 + 11).
Input: N = 10
Output: -1
Explanation:
It is not possible to construct array of size 10.
Approach: To solve the problem mentioned above the very first observation is that it is not possible to create an array that has size N which is a multiple of 2 but not multiple of 4. Because, if that happens then the sum of one half which contains odd numbers will always be odd and the sum of another half which contains even numbers will always be even, hence the sum of both halves can’t be the same.
Therefore, the array which satisfies the problem statement should always have a size N which is a multiple of 4. The approach is to first construct N/2 even numbers starting from 2, which is the first half of the array. Then create another part of the array starting from 1 and finally calculate the last odd element such that it makes both the halves equal. In order to do so the last element of odd numbers should be (N/2) – 1 + N.
Below is the implementation of the above approach:
CPP
// C++ program to Create an array// of size N consisting of distinct// elements where sum of odd elements// is equal to sum of even elements#include <bits/stdc++.h>using namespace std;// Function to construct the required arrayvoid arrayConstruct(int N){ // To construct first half, // distinct even numbers for (int i = 2; i <= N; i = i + 2) cout << i << " "; // To construct second half, // distinct odd numbers for (int i = 1; i < N - 1; i = i + 2) cout << i << " "; // Calculate the last number of second half // so as to make both the halves equal cout << N - 1 + (N / 2) << endl;}// Function to construct the required arrayvoid createArray(int N){ // check if size is multiple of 4 // then array exist if (N % 4 == 0) // function call to construct array arrayConstruct(N); else cout << -1 << endl;}// Driver codeint main(){ int N = 8; createArray(N); return 0;} |
Java
// Java program to Create an array// of size N consisting of distinct// elements where sum of odd elements// is equal to sum of even elementsclass GFG{ // Function to construct the required arraystatic void arrayConstruct(int N){ // To confirst half, // distinct even numbers for (int i = 2; i <= N; i = i + 2) System.out.print(i+ " "); // To consecond half, // distinct odd numbers for (int i = 1; i < N - 1; i = i + 2) System.out.print(i+ " "); // Calculate the last number of second half // so as to make both the halves equal System.out.print(N - 1 + (N / 2) +"\n");} // Function to construct the required arraystatic void createArray(int N){ // check if size is multiple of 4 // then array exist if (N % 4 == 0) // function call to conarray arrayConstruct(N); else System.out.print(-1 +"\n");} // Driver codepublic static void main(String[] args){ int N = 8; createArray(N);}}// This code is contributed by Princi Singh |
Python3
# python3 program to Create an array# of size N consisting of distinct# elements where sum of odd elements# is equal to sum of even elements# Function to construct the required arraydef arrayConstruct(N): # To construct first half, # distinct even numbers for i in range(2, N + 1, 2): print(i,end=" ") # To construct second half, # distinct odd numbers for i in range(1, N - 1, 2): print(i, end=" ") # Calculate the last number of second half # so as to make both the halves equal print(N - 1 + (N // 2))# Function to construct the required arraydef createArray(N): # check if size is multiple of 4 # then array exist if (N % 4 == 0): # function call to construct array arrayConstruct(N) else: cout << -1 << endl# Driver codeif __name__ == '__main__': N = 8 createArray(N)# This code is contributed by mohit kumar 29 |
C#
// C# program to Create an array// of size N consisting of distinct// elements where sum of odd elements// is equal to sum of even elementsusing System;class GFG{ // Function to construct the required arraystatic void arrayConstruct(int N){ // To confirst half, // distinct even numbers for (int i = 2; i <= N; i = i + 2) Console.Write(i + " "); // To consecond half, // distinct odd numbers for (int i = 1; i < N - 1; i = i + 2) Console.Write(i + " "); // Calculate the last number of second half // so as to make both the halves equal Console.Write(N - 1 + (N / 2) +"\n");} // Function to construct the required arraystatic void createArray(int N){ // check if size is multiple of 4 // then array exist if (N % 4 == 0) // function call to conarray arrayConstruct(N); else Console.Write(-1 +"\n");} // Driver codepublic static void Main(String[] args){ int N = 8; createArray(N);}}// This code is contributed by Rajput-Ji |
Javascript
<script>// JavaScript program to Create an array// of size N consisting of distinct// elements where sum of odd elements// is equal to sum of even elements// Function to construct the required arrayfunction arrayConstruct(N){ // To construct first half, // distinct even numbers for (let i = 2; i <= N; i = i + 2) document.write(i + " "); // To construct second half, // distinct odd numbers for (let i = 1; i < N - 1; i = i + 2) document.write(i + " "); // Calculate the last number of second half // so as to make both the halves equal document.write(N - 1 + (N / 2) + "<br>");}// Function to construct the required arrayfunction createArray(N){ // check if size is multiple of 4 // then array exist if (N % 4 == 0) // function call to construct array arrayConstruct(N); else document.write(-1 + "<br>");}// Driver code let N = 8; createArray(N);// This code is contributed by Surbhi Tyagi.</script> |
2 4 6 8 1 3 5 11
Time Complexity: O(N)
Auxiliary Space: O(1)
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