# Number of ways to divide a N elements equally into group of at least 2

Given an integer **N** denoting the number of elements, the task is to find the number of ways to divide these elements equally into groups such that each group has at least 2 elements.

**Examples:**

Input:N = 2Output:1Explanation:There can be only one group.

Input:N = 10Output:3Explanation:There are 3 ways to divide elements:

One group having all the 10 elements.

Two groups where each group has 5 elements.

Five groups where each group has 2 elements.

**Approach:** The above problem can be solved using the below-given brute force approach. In every iteration of the loop, **i*** *represents the number of groups. If **N*** *is completely* *divisible by **i**, hence, the elements can be equally divided among groups. Follow the steps below to solve the problem:

- Declare variable
and initialize it by 0.**ways** - Iterate over the range
**[1, N/2]**using the variable**i**and perform the following tasks:- Check if
**N**is completely divisible by**i**. - If yes, then increment
by**ways****1**.

- Check if
- After performing the above steps, print the value of
**ways**as the answer.

Below is the implementation of the above approach:

## C++

`// C++ program for the given approach` `#include <iostream>` `using` `namespace` `std;` `// Function to find the number of ways` `int` `numberofWays(` `int` `N)` `{` ` ` `// Variable to store the number of ways` ` ` `int` `ways = 0;` ` ` `int` `i;` ` ` `// Loop to find total number of ways` ` ` `for` `(i = 1; i <= N / 2; i++) {` ` ` `if` `(N % i == 0)` ` ` `ways++;` ` ` `}` ` ` `// Returning the number of ways` ` ` `return` `ways;` `}` `// Driver Code` `int` `main()` `{` ` ` `// Declaring and initialising N` ` ` `int` `N = 10;` ` ` `// Function call` ` ` `int` `ans = numberofWays(N);` ` ` `// Displaying the answer on screen` ` ` `cout << ans;` ` ` `return` `0;` `}` |

## Java

`// Java program for the above approach` `import` `java.io.*;` `import` `java.lang.*;` `import` `java.util.*;` `class` `GFG {` ` ` `// Function to find the number of ways` ` ` `static` `int` `numberofWays(` `int` `N)` ` ` `{` ` ` `// Variable to store the number of ways` ` ` `int` `ways = ` `0` `;` ` ` `int` `i;` ` ` `// Loop to find total number of ways` ` ` `for` `(i = ` `1` `; i <= N / ` `2` `; i++) {` ` ` `if` `(N % i == ` `0` `)` ` ` `ways++;` ` ` `}` ` ` `// Returning the number of ways` ` ` `return` `ways;` ` ` `}` ` ` `public` `static` `void` `main (String[] args)` ` ` `{` ` ` `// Declaring and initialising N` ` ` `int` `N = ` `10` `;` ` ` `// Function call` ` ` `int` `ans = numberofWays(N);` ` ` `// Displaying the answer on screen` ` ` `System.out.print(ans);` ` ` `}` `}` `// This code is contributed by hrithikgarg03188` |

## Python3

`# Python code for the above approach` `# Function to find the number of ways` `def` `numberofWays(N):` ` ` `# Variable to store the number of ways` ` ` `ways ` `=` `0` `;` ` ` `i ` `=` `None` ` ` `# Loop to find total number of ways` ` ` `for` `i ` `in` `range` `(` `1` `, (N ` `/` `/` `2` `) ` `+` `1` `):` ` ` `if` `(N ` `%` `i ` `=` `=` `0` `):` ` ` `ways ` `+` `=` `1` ` ` `# Returning the number of ways` ` ` `return` `ways;` `# Driver Code` `# Declaring and initialising N` `N ` `=` `10` `;` `# Function call` `ans ` `=` `numberofWays(N);` `# Displaying the answer on screen` `print` `(ans);` `# This code is contributed by Saurabh Jaiswal` |

## Javascript

`<script>` ` ` `// JavaScript code for the above approach` ` ` `// Function to find the number of ways` ` ` `function` `numberofWays(N)` ` ` `{` ` ` ` ` `// Variable to store the number of ways` ` ` `let ways = 0;` ` ` `let i;` ` ` `// Loop to find total number of ways` ` ` `for` `(i = 1; i <= Math.floor(N / 2); i++) {` ` ` `if` `(N % i == 0)` ` ` `ways++;` ` ` `}` ` ` `// Returning the number of ways` ` ` `return` `ways;` ` ` `}` ` ` `// Driver Code` ` ` `// Declaring and initialising N` ` ` `let N = 10;` ` ` `// Function call` ` ` `let ans = numberofWays(N);` ` ` `// Displaying the answer on screen` ` ` `document.write(ans);` ` ` `// This code is contributed by Potta Lokesh` ` ` `</script>` |

**Output**

3

* Time Complexity:* O(N)

*O(1)*

**Auxiliary Space:**