# 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 = 12Output:3Explanation:

The integer N = 12 can be divided into 3 consecutive elements {3, 4, 5} where 3 + 4 + 5 = 12.Input:N = 8Output:-1Explanation:

No such division of integer 8 is possible.Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the

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**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:

- Iterate from index(say
**i**) from 2. - Find the sum of first i numbers(say
**sum**). - For any iteration if
**(N – sum)**is divisible by i then print that value of**i**. - 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 <bits/stdc++.h>` `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` |

## Javascript

`<script>` ` ` `// Javascript program for the above approach` ` ` ` ` `// Function to find the K consecutive` ` ` `// elements with a sum equal to N` ` ` `function` `canBreakN(n)` ` ` `{` ` ` `// Iterate over [2, INF]` ` ` `for` `(let i = 2;; i++)` ` ` `{` ` ` `// Store the sum` ` ` `let m = parseInt(i * (i + 1) / 2, 10);` ` ` `// If the sum exceeds N` ` ` `// then break the loop` ` ` `if` `(m > n)` ` ` `break` `;` ` ` `let k = n - m;` ` ` `// Common difference should be` ` ` `// divisible by number of terms` ` ` `if` `(k % i != 0)` ` ` `continue` `;` ` ` `// Print value of i & return` ` ` `document.write(i);` ` ` `return` `;` ` ` `}` ` ` `// Print "-1" if not possible` ` ` `// to break N` ` ` `document.write(` `"-1"` `);` ` ` `}` ` ` ` ` `// Given N` ` ` `let N = 12;` ` ` ` ` `// Function call` ` ` `canBreakN(N);` `// This code is contributed by sureh07.` `</script>` |

**Output:**

3

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