# Check if N can be divided into K consecutive elements with a sum equal to N

Given an integer N, our task is to check if N can be divided into K consecutive elements with a sum equal to N. Print -1 if it is not possible to divide in this manner, otherwise print the value K.
Examples:

Input: N = 12
Output:
Explanation:
The integer N = 12 can be divided into 3 consecutive elements {3, 4, 5} where 3 + 4 + 5 = 12.

Input: N = 8
Output: -1
Explanation:
No such division of integer 8 is possible.

Approach: To solve the problem mentioned above let us divide the integer N into i consecutive numbers. The terms of the sequence will look like (d+1), (d+2), (d+3)…..(d+i) where d is the common difference present in each of the integers and the sum of this sequence should be equal to N
So, the sum of these number can also be expressed as: As the sum = i * (i + 1) / 2 grows quadratically, we have, N – sum = i * d. Hence, for a solution to exist, the number of integers should evenly divide the quantity N – sum. Below are the steps:

1. Iterate from index(say i) from 2.
2. Find the sum of first i numbers(say sum).
3. For any iteration if (N – sum) is divisible by i then print that value of i.
4. For any iteration if N exceeds the sum then print “-1”.

Below is the implementation of the above approach:

## C++

 // C++ program for the above approach  #include  using namespace std;     // Function to find the K consecutive  // elements with a sum equal to N  void canBreakN(long long n)  {      // Iterate over [2, INF]      for (long long i = 2;; i++) {             // Store the sum          long long m = i * (i + 1) / 2;             // If the sum exceeds N          // then break the loop          if (m > n)              break;             long long k = n - m;             // Common difference should be          // divisible by number of terms          if (k % i)              continue;             // Print value of i & return          cout << i << endl;          return;      }         // Print "-1" if not possible      // to break N      cout << "-1";  }     // Driver Code  int main()  {      // Given N      long long N = 12;         // Function Call      canBreakN(N);      return 0;  }

## Java

 // Java program for the above approach  class GFG{     // Function to find the K consecutive  // elements with a sum equal to N  public static void canBreakN(long n)  {             // Iterate over [2, INF]      for(long i = 2;; i++)       {                     // Store the sum          long m = i * (i + 1) / 2;             // If the sum exceeds N          // then break the loop          if (m > n)              break;             long k = n - m;             // Common difference should be          // divisible by number of terms          if (k % i != 0)              continue;             // Print value of i & return          System.out.println(i);          return;      }         // Print "-1" if not possible      // to break N      System.out.println("-1");  }     // Driver Code  public static void main(String[] args)  {             // Given N      long N = 12;         // Function call      canBreakN(N);  }  }     // This code is contributed by jrishabh99

## Python3

 # Python3 program for the above approach     # Function to find the K consecutive  # elements with a sum equal to N  def canBreakN(n):             # Iterate over [2, INF]      for i in range(2, n):             # Store the sum          m = i * (i + 1) // 2            # If the sum exceeds N          # then break the loop          if (m > n):              break            k = n - m             # Common difference should be          # divisible by number of terms          if (k % i):              continue            # Print value of i & return          print(i)          return            # Print "-1" if not possible      # to break N      print("-1")     # Driver Code     # Given N  N = 12    # Function call  canBreakN(N)     # This code is contributed by code_hunt

## C#

 // C# program for the above approach  using System;  class GFG{      // Function to find the K consecutive  // elements with a sum equal to N  public static void canBreakN(long n)  {              // Iterate over [2, INF]      for(long i = 2;; i++)       {                      // Store the sum          long m = i * (i + 1) / 2;              // If the sum exceeds N          // then break the loop          if (m > n)              break;              long k = n - m;              // Common difference should be          // divisible by number of terms          if (k % i != 0)              continue;              // Print value of i & return          Console.Write(i);          return;      }          // Print "-1" if not possible      // to break N      Console.Write("-1");  }      // Driver Code  public static void Main(string[] args)  {              // Given N      long N = 12;          // Function call      canBreakN(N);  }  }      // This code is contributed by rock_cool

Output:

3


Time Complexity: O(K), where K is the number of element whose sum is K.
Auxiliary Space: O(1)

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