Find N distinct integers with sum N

Namachila November 19, 2025

0 0 4 minutes read

Given an integer N, the task is to find N distinct integers whose sum is N. If there is more than one combination of the integers, print any one of them.

Examples:

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

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

Approach: The idea is to print N/2 Symmetric Pairs like (+x, -x) so that the resultant sum will always be 0.
Now if integer N is odd, then print N along with these set of integers to make sum of all integers equals to N
If N is even, print 0 and N along with these set of integers to make sum of all integers equals to N.

Below is the implementation of the above approach:

C++

// C++ for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to print distinct N
// numbers whose sum is N
void findNumbers(int N)
{
    // To store how many symmetric
    // pairs needs to be calculated
    int half = N / 2;
 
    // For even N we have to print
    // one less symmetric pair
    if (N % 2 == 0) {
        half--;
    }
 
    // Iterate till [1 n/2] and Print
    // all symmetric pairs(i, -i)
    for (int i = 1; i <= half; i++) {
 
        // Print 2 symmetric numbers
        cout << (-1) * i
             << ", " << i << ", ";
    }
 
    // if N is Odd, then print N
    if (N & 1) {
        cout << N << endl;
    }
 
    // Else print(0, N)
  else {
    cout << 0 << ", "
         << N << endl;
   }
}
 
// Driver Code
int main()
{
    // Given Sum
    int N = 5;
 
    // Function Call
    findNumbers(N);
    return 0;
}

Java

// Java for the above approach
class GFG{
     
// Function to print distinct N 
// numbers whose sum is N 
public static void findNumbers(int N) 
{ 
     
    // To store how many symmetric 
    // pairs needs to be calculated 
    int half = N / 2; 
     
    // For even N we have to print 
    // one less symmetric pair 
    if (N % 2 == 0)
    { 
        half--; 
    } 
     
    // Iterate till [1 n/2] and Print 
    // all symmetric pairs(i, -i) 
    for(int i = 1; i <= half; i++)
    { 
 
       // Print 2 symmetric numbers 
       System.out.print((-1) * i + ", " +
                               i + ", ");
    } 
     
    // if N is Odd, then print N 
    int check = N & 1;
    if (check != 0)
    { 
        System.out.println(N);
    } 
     
    // Else print(0, N)
    else
    {
    System.out.println(0 + ", " + N);
    }
} 
 
// Driver code
public static void main(String[] args)
{
         
    // Given sum 
    int N = 5; 
     
    // Function sall 
    findNumbers(N); 
}
}
 
// This code is contributed by divyeshrabadiya07        

Python3

# Python3 code for the above approach 
 
# Function to print distinct N 
# numbers whose sum is N 
def findNumbers(N):
 
    # To store how many symmetric 
    # pairs needs to be calculated 
    half = int(N / 2) 
 
    # For even N we have to print 
    # one less symmetric pair 
    if (N % 2 == 0): 
        half = half - 1
 
    # Iterate till [1 n/2] and Print 
    # all symmetric pairs(i, -i) 
    for i in range(1, half + 1):
 
        # Print 2 symmetric numbers 
        print((-1) * i, end = ', ')
        print(i, end = ', ')
 
    # If N is Odd, then print N 
    if (N & 1): 
        print(N, end = '\n')
 
    # Else print(0, N)
    else:
        print(0, end = ', ')
        print(N, end = '\n') 
 
# Driver Code 
N = 5
 
# Function Call 
findNumbers(N)
 
# This code is contributed by PratikBasu    

C#

// C# for the above approach
using System;
class GFG{
     
// Function to print distinct N 
// numbers whose sum is N 
public static void findNumbers(int N) 
{ 
     
    // To store how many symmetric 
    // pairs needs to be calculated 
    int half = N / 2; 
     
    // For even N we have to print 
    // one less symmetric pair 
    if (N % 2 == 0)
    { 
        half--; 
    } 
     
    // Iterate till [1 n/2] and Print 
    // all symmetric pairs(i, -i) 
    for(int i = 1; i <= half; i++)
    { 
 
        // Print 2 symmetric numbers 
        Console.Write((-1) * i + ", " +
                             i + ", ");
    } 
     
    // if N is Odd, then print N 
    int check = N & 1;
    if (check != 0)
    { 
        Console.Write(N + "\n");
    } 
     
    // Else print(0, N)
    else
    {
    Console.Write(0 + ", " + N + "\n");
    }
} 
 
// Driver code
public static void Main(string[] args)
{
         
    // Given sum 
    int N = 5; 
     
    // Function sall 
    findNumbers(N); 
}
}
 
// This code is contributed by rutvik_56

Javascript

<script>
 
// javascript program for the above approach
 
// Function to print distinct N
// numbers whose sum is N
function findNumbers( N)
{
    // To store how many symmetric
    // pairs needs to be calculated
    let half = parseInt(N / 2);
 
    // For even N we have to print
    // one less symmetric pair
    if (N % 2 == 0) {
        half--;
    }
 
    // Iterate till [1 n/2] and Print
    // all symmetric pairs(i, -i)
    for (let i = 1; i <= half; i++) {
 
        // Print 2 symmetric numbers
         document.write( (-1) * i
             + ", " + i + ", ");
    }
 
    // if N is Odd, then print N
    if (N & 1) {
         document.write( N);
    }
 
    // Else print(0, N)
  else {
     document.write(  0 + ", "
         + N +"<br/>");
   }
}
 
// Driver Code
 
    // Given Sum
    let N = 5;
 
    // Function Call
    findNumbers(N);
           
 
    // This code contributed by aashish1995 
 
</script>

Output:

-1,1,-2,2,5

Time Complexity: O(N/2) which is asymptotically same as O(N).

Space Complexity: O(1) as no extra space has been used.

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