Given a **positive integer N**, count number of ways to write N as a sum of three numbers. For numbers which are not expressible print -1.**Examples:**

Input:N = 4Output:3Explanation:

( 1 + 1 + 2 ) = 4

( 1 + 2 + 1 ) = 4

( 2 + 1 + 1 ) = 4.

So in total, there are 3 ways.Input:N = 5Output:6

( 1 + 1 + 3 ) = 5

( 1 + 3 + 1 ) = 5

( 3 + 1 + 1 ) = 5

( 1 + 2 + 2 ) = 5

( 2 + 2 + 1 ) = 5

( 2 + 1 + 2 ) = 5.

So in total, there are 6 ways

**Approach:** To solve the problem mentioned above if we take a closer look we will observe a pattern in solution to the question. For all the numbers that are greater than 2 we get a series 3, 6, 10, 15, 25 and so on, which is nothing but** the ****sum of first N-1 natural numbers****.**

Below is the implementation of the above approach:

## C++

`// C++ program to count the total number of` `// ways to write N as a sum of three numbers` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to find the number of ways` `void` `countWays(` `int` `n)` `{` ` ` `// Check if number is less than 2` ` ` `if` `(n <= 2)` ` ` `cout << ` `"-1"` `;` ` ` `else` `{` ` ` `// Calculate the sum` ` ` `int` `ans = (n - 1) * (n - 2) / 2;` ` ` `cout << ans;` ` ` `}` `}` `// Driver code` `int` `main()` `{` ` ` `int` `N = 5;` ` ` `countWays(N);` ` ` `return` `0;` `}` |

## Java

`// Java program to count the total number of` `// ways to write N as a sum of three numbers` `class` `GFG{` `// Function to find the number of ways` `static` `void` `countWays(` `int` `n)` `{` ` ` ` ` `// Check if number is less than 2` ` ` `if` `(n <= ` `2` `)` ` ` `{` ` ` `System.out.print(` `"-1"` `);` ` ` `}` ` ` `else` ` ` `{` ` ` ` ` `// Calculate the sum` ` ` `int` `ans = (n - ` `1` `) * (n - ` `2` `) / ` `2` `;` ` ` `System.out.print(ans);` ` ` `}` `}` `// Driver code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `N = ` `5` `;` ` ` `countWays(N);` `}` `}` `// This code is contributed by Amit Katiyar` |

## Python3

`# Python3 program to count the total number of` `# ways to write N as a sum of three numbers` `def` `countWays(N):` ` ` `# Check if number is less than 2` ` ` `if` `(N <` `=` `2` `):` ` ` `print` `(` `"-1"` `)` ` ` `else` `:` ` ` ` ` `# Calculate the sum` ` ` `ans ` `=` `(N ` `-` `1` `) ` `*` `(N ` `-` `2` `) ` `/` `2` ` ` ` ` `print` `(ans)` `# Driver code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` ` ` `N ` `=` `5` ` ` `countWays(N)` `# This code is contributed by coder001` |

## C#

`// C# program to count the total number of` `// ways to write N as a sum of three numbers` `using` `System;` `class` `GFG{` ` ` `// Function to find the number of ways` `static` `void` `countWays(` `int` `n)` `{` ` ` ` ` `// Check if number is less than 2` ` ` `if` `(n <= 2)` ` ` `{` ` ` `Console.WriteLine(` `"-1"` `);` ` ` `}` ` ` `else` ` ` `{` ` ` `// Calculate the sum` ` ` `int` `ans = (n - 1) * (n - 2) / 2;` ` ` `Console.WriteLine(ans);` ` ` `}` `}` `// Driver code ` `static` `void` `Main()` `{` ` ` `int` `N = 5;` ` ` `countWays(N);` `}` `}` `// This code is contributed by divyeshrabadiya07 ` |

## Javascript

`<script>` `// Javascript program to count the total number of` `// ways to write N as a sum of three numbers` `// Function to find the number of ways` `function` `countWays(n)` `{` ` ` `// Check if number is less than 2` ` ` `if` `(n <= 2)` ` ` `document.write( ` `"-1"` `);` ` ` `else` `{` ` ` `// Calculate the sum` ` ` `var` `ans = (n - 1) * (n - 2) / 2;` ` ` `document.write( ans);` ` ` `}` `}` `// Driver code` `var` `N = 5;` `countWays(N);` `</script>` |

**Output:**

6

**Time Complexity:** O(1)

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