Find N distinct integers with zero sum

Given an integer N, our task is to print N distinct numbers such that their sum is 0.
Examples:

Input: N = 3 
Output: 1, -1, 0 
Explanation:
On adding the numbers that is 1 + (-1) + 0 the sum is 0.
Input: N = 4 
Output: 1, -1, 2, -2 
Explanation:
On adding the numbers that is 1 + (-1) + 2 + (-2) the sum is 0. 

Approach: To solve the problem mentioned above the main idea is to print Symmetric Pairs like (+x, -x) so that the sum will always be 0. The edge case to the problem is to observe that if integer N is odd, then print one 0 along with the numbers so that sum is not affected.
Below is the implementation of the above approach: 

C++

// C++ implementation to Print N distinct
// numbers such that their sum is 0

#include <bits/stdc++.h>
using namespace std;

// Function to print distinct n
// numbers such that their sum is 0
void findNumbers(int N)
{
    for (int i = 1; i <= N / 2; i++) {

        // Print 2 symmetric numbers
        cout << i << ", " << -i << ", ";
    }

    // print a extra 0 if N is odd
    if (N % 2 == 1)
        cout << 0;
}

// Driver code
int main()
{
    int N = 10;

    findNumbers(N);
}

Java

// Java implementation to Print N distinct
// numbers such that their sum is 0

class GFG{

// Function to print distinct n
// numbers such that their sum is 0
static void findNumbers(int N)
{
    for (int i = 1; i <= N / 2; i++)
    {
        // Print 2 symmetric numbers
        System.out.print(i + ", " + -i + ", ");
    }

    // Print a extra 0 if N is odd
    if (N % 2 == 1)
        System.out.print(0);
}

// Driver code
public static void main(String[] args)
{
    int N = 10;
    findNumbers(N);
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 implementation to print N distinct
# numbers such that their sum is 0

# Function to print distinct n
# numbers such that their sum is 0
def findNumbers(N):

    for i in range(1, N // 2 + 1):

        # Print 2 symmetric numbers
        print(i, end = ', ')
        print(-i, end = ', ')

    # Print a extra 0 if N is odd
    if N % 2 == 1:
        print(0, end = '')

# Driver code
if __name__=='__main__':

    N = 10
    findNumbers(N)

# This code is contributed by rutvik_56

C#

// C# implementation to print N distinct
// numbers such that their sum is 0
using System;

class GFG {

// Function to print distinct n
// numbers such that their sum is 0
static void findNumbers(int N)
{
    for(int i = 1; i <= (N / 2); i++)
    {

       // Print 2 symmetric numbers
       Console.Write(i + ", " + -i + ", ");
    }

    // Print a extra 0 if N is odd
    if (N % 2 == 1)
        Console.Write(0);
}

// Driver code
static void Main()
{
    int N = 10;

    findNumbers(N);
}
}

// This code is contributed by divyeshrabadiya07

Javascript

<script>

// Javascript implementation to Print N distinct
// numbers such that their sum is 0

// Function to print distinct n
// numbers such that their sum is 0
function findNumbers(N)
{
    for (var i = 1; i <= N / 2; i++) {

        // Print 2 symmetric numbers
        document.write( i + ", " + -i + ", ");
    }

    // print a extra 0 if N is odd
    if (N % 2 == 1)
        document.write( 0);
}

// Driver code
var N = 10;
findNumbers(N);

</script>

Output:

1, -1, 2, -2, 3, -3, 4, -4, 5, -5,

Time Complexity: O(N)

Auxiliary Space: O(1)