# Check whether N items can be divided into K groups of unique size

Given integers **N** and **K**, the task is to check if it is possible to divide **N** numbers into **K** groups such that all the **K** groups are of different size and each part has at least one number.

**Examples:**

Input:N = 5, K = 2Output:YesExplanation:5 numbers can be divided into 2 parts of different size. The possible size of the groups can be (1, 4) and (2, 3).

Input:N = 3, K = 3Output:NoExplanation:3 numbers cannot be divide into 3 groups of unique size.

**Approach:** The problem can be solved on the basis of following observation.

To divide N numbers into K groups such that each group has at least one number and no two groups have same size:

- There should be at least K numbers. If N < K, then division is not possible.
- If N > K then the K groups will be at least of size 1, 2, 3, 4 . . . K . So N must be at least K*(K + 1)/2.

Therefore, the condition to be satisfied isN ≥ K*(K + 1)/2.

Below is the implementation of the above approach.

## C++

`// C++ code to implement above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to check if it is possible` `// to break N in K groups` `void` `checkPartition(` `int` `N, ` `int` `K)` `{` ` ` `// Invalid case` ` ` `if` `(N < (K * (K + 1)) / 2) {` ` ` `cout << ` `"No"` `;` ` ` `}` ` ` `else` `{` ` ` `cout << ` `"Yes"` `;` ` ` `}` `}` `// Driver code` `int` `main()` `{` ` ` `int` `N = 6, K = 5;` ` ` `checkPartition(N, K);` ` ` `return` `0;` `}` |

## C

`// C code to implement above approach` `#include <stdio.h>` `// Function to check if it is possible` `// to break N in K groups` `void` `checkPartition(` `int` `N, ` `int` `K)` `{` ` ` `// Invalid case` ` ` `if` `(N < (K * (K + 1)) / 2) {` ` ` `printf` `(` `"No"` `);` ` ` `}` ` ` `else` `{` ` ` `printf` `(` `"Yes"` `);` ` ` `}` `}` `// Driver code` `int` `main()` `{` ` ` `int` `N = 6, K = 5;` ` ` `checkPartition(N, K);` ` ` `return` `0;` `}` |

## Java

`// Java code to implement above approach` `import` `java.io.*;` `class` `GFG {` ` ` `// Function to check if it is possible` ` ` `// to break N in K groups` ` ` `public` `static` `void` `checkPartition(` `int` `N, ` `int` `K)` ` ` `{` ` ` `// Invalid case` ` ` `if` `(N < (K * (K + ` `1` `) / ` `2` `)) {` ` ` `System.out.print(` `"No"` `);` ` ` `}` ` ` `else` `{` ` ` `System.out.print(` `"Yes"` `);` ` ` `}` ` ` `}` ` ` `// Driver code` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `int` `N = ` `6` `, K = ` `5` `;` ` ` `checkPartition(N, K);` ` ` `}` `}` |

## Python3

`# Python code to implement above approach` `def` `checkPartition(N, K):` ` ` `# Invalid case` ` ` `if` `(N < (K` `*` `(K ` `+` `1` `))` `/` `/` `2` `):` ` ` `print` `(` `"No"` `)` ` ` `else` `:` ` ` `print` `(` `"Yes"` `)` `# Driver code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` `N ` `=` `6` ` ` `K ` `=` `5` ` ` `checkPartition(N, K)` |

## C#

`// C# code to implement above approach` `using` `System;` `class` `GFG {` ` ` `// Function to check if it is possible` ` ` `// to break N in K groups` ` ` `public` `static` `void` `checkPartition(` `int` `N, ` `int` `K)` ` ` `{` ` ` `// Invalid case` ` ` `if` `(N < (K * (K + 1) / 2)) {` ` ` `Console.Write(` `"No"` `);` ` ` `}` ` ` `else` `{` ` ` `Console.Write(` `"Yes"` `);` ` ` `}` ` ` `}` ` ` `// Driver code` ` ` `public` `static` `void` `Main()` ` ` `{` ` ` `int` `N = 6, K = 5;` ` ` `checkPartition(N, K);` ` ` `}` `}` `// This code is contributed by saurabh_jaiswal.` |

## Javascript

`<script>` ` ` `// JavaScript code for the above approach` ` ` `// Function to check if it is possible` ` ` `// to break N in K groups` ` ` `function` `checkPartition(N, K)` ` ` `{` ` ` ` ` `// Invalid case` ` ` `if` `(N < Math.floor((K * (K + 1)) / 2)) {` ` ` `document.write(` `"No"` `);` ` ` `}` ` ` `else` `{` ` ` `document.write(` `"Yes"` `);` ` ` `}` ` ` `}` ` ` `// Driver code` ` ` `let N = 6, K = 5;` ` ` `checkPartition(N, K);` ` ` `// This code is contributed by Potta Lokesh` ` ` `</script>` |

**Output**

No

* Time Complexity:* O(1)

*O(1).*

**Auxiliary Space:**