# Find K consecutive integers such that their sum is N

Given two integers N and K, the task is to find K consecutive integers such that their sum if N.

Note: If there is no such K integers print -1.
Examples:

Input: N = 15, K = 5
Output: 1 2 3 4 5
Explanation:
N can be represented as sum of 5 consecutive integers as follows –
=> N => 1 + 2 + 3 + 4 + 5 = 15

Input: N = 33, K = 6
Output: 3 4 5 6 7 8
Explanation:
N can be represented as sum of 6 consecutive integers as follows –
=> N => 3 + 4 + 5 + 6 + 7 + 8 = 33

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive Approach: A simple solution is to run a loop from i = 0 to N – (K – 1) to check if K consecutive integers starting from i is having sum as N.

Efficient Approach: The idea is to use Arithmetic Progression to solve this problem, where sum of K terms of arithmetic progression with common difference is 1 can be defined as follows –

1. Sum of K Terms – => 2. Solving the equation further to get the first term possible
=> 3. Here aK is the Kth term which can be written as a1 + K – 1
=> => 4. Finally, check the first term computed is an integer, If yes then K consecutive number exists whose sum if N.

Below is the implementation of the above approach:

## C++

 // C++ implementation to check if  // a number can be expressed as  // sum of K consecutive integer     #include  using namespace std;     // Function to check if a number can be  // expressed as the sum of k consecutive  void checksum(int n, int k)  {      // Finding the first      // term of AP      float first_term = ((2 * n) / k                          + (1 - k))                         / 2.0;         // Checking if first      // term is an integer      if (first_term - int(first_term) == 0) {             // Loop to print the K          // consecutive integers          for (int i = first_term;               i <= first_term + k - 1; i++) {              cout << i << " ";          }      }      else         cout << "-1";  }     // Driver Code  int main()  {      int n = 33, k = 6;      checksum(n, k);      return 0;  }

## Java

 // Java implementation to check if  // a number can be expressed as  // sum of K consecutive integer  class GFG{     // Function to check if a number can be  // expressed as the sum of k consecutive  static void checksum(int n, int k)  {             // Finding the first      // term of AP      float first_term = (float) (((2 * n) / k +                                    (1 - k)) / 2.0);         // Checking if first      // term is an integer      if (first_term - (int)(first_term) == 0)      {             // Loop to print the K          // consecutive integers          for(int i = (int)first_term;               i <= first_term + k - 1; i++)          {             System.out.print(i + " ");          }      }      else         System.out.print("-1");  }     // Driver Code  public static void main(String[] args)  {      int n = 33, k = 6;             checksum(n, k);  }  }     // This code is contributed by 29AjayKumar

## Python3

 # Python3 implementation to check    # if a number can be expressed as   # sum of K consecutive integer      # Function to check if a number can be   # expressed as the sum of k consecutive   def checksum(n, k):             # Finding the first       # term of AP       first_term = ((2 * n) / k + (1 - k)) / 2.0            # Checking if first       # term is an integer       if (first_term - int(first_term) == 0):                     # Loop to print the K           # consecutive integers           for i in range(int(first_term),                         int(first_term) + k):              print(i, end = ' ')      else:          print('-1')     # Driver Code   if __name__=='__main__':             (n, k) = (33, 6)      checksum(n, k)     # This code is contributed by rutvik_56

## C#

 // C# implementation to check if  // a number can be expressed as  // sum of K consecutive integer  using System;  class GFG{     // Function to check if a number can be  // expressed as the sum of k consecutive  static void checksum(int n, int k)  {             // Finding the first      // term of AP      float first_term = (float)(((2 * n) / k +                                   (1 - k)) / 2.0);         // Checking if first      // term is an integer      if (first_term - (int)(first_term) == 0)      {             // Loop to print the K          // consecutive integers          for(int i = (int)first_term;                   i <= first_term + k - 1; i++)          {              Console.Write(i + " ");          }      }      else         Console.Write("-1");  }     // Driver Code  public static void Main(String[] args)  {      int n = 33, k = 6;             checksum(n, k);  }  }     // This code is contributed by sapnasingh4991

Output:

3 4 5 6 7 8


Don’t stop now and take your learning to the next level. Learn all the important concepts of Data Structures and Algorithms with the help of the most trusted course: DSA Self Paced. Become industry ready at a student-friendly price.

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :
Practice Tags :

1

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.