# Maximize number of groups formed with size not smaller than its largest element

Given an array arr[] of N positive integers(1 ≤ arr[i] ≤ N ), divide the elements of the array into groups such that the size of each group is greater than or equal to the largest element of that group. It may be also possible that an element cannot join any group. The task is to maximize the number of groups.

Examples:

Input: arr = {2, 3, 1, 2, 2}
Output: 2
Explanation:
In the first group we can take {1, 2}
In the second group we can take {2, 2, 3}
Therefore the maximum 2 groups can be possible.

Input: arr = {1, 1, 1}
Output: 3

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

• Firstly store the number of occurrence of each element in an array.
• Now, Make groups of similar elements. For example: if there are three 1s in the array then make three groups for each 1.
• Then store remaining elements and start grouping from the lowest element.

Below is the implementation of the above approach.

## C++

 `// C++ implementation of above approach ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function that prints the number ` `// of maximum groups ` `void` `makeGroups(``int` `a[], ``int` `n) ` `{ ` `    ``vector<``int``> v(n + 1, 0); ` ` `  `    ``// Store the number of ` `    ``// occurrence of elements ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``v[a[i]]++; ` `    ``} ` ` `  `    ``int` `no_of_groups = 0; ` ` `  `    ``// Make all groups of similar ` `    ``// elements and store the ` `    ``// left numbers ` `    ``for` `(``int` `i = 1; i <= n; i++) { ` `        ``no_of_groups += v[i] / i; ` ` `  `        ``v[i] = v[i] % i; ` `    ``} ` ` `  `    ``int` `i = 1; ` `    ``int` `total = 0; ` ` `  `    ``for` `(i = 1; i <= n; i++) { ` `        ``// Condition for finding first ` `        ``// leftover element ` `        ``if` `(v[i] != 0) { ` `            ``total = v[i]; ` `            ``break``; ` `        ``} ` `    ``} ` ` `  `    ``i++; ` ` `  `    ``while` `(i <= n) { ` `        ``// Condition for current ` `        ``// leftover element ` `        ``if` `(v[i] != 0) { ` `            ``total += v[i]; ` ` `  `            ``// Condition if group size ` `            ``// is equal to or more than ` `            ``// current element ` `            ``if` `(total >= i) { ` `                ``int` `rem = total - i; ` `                ``no_of_groups++; ` `                ``total = rem; ` `            ``} ` `        ``} ` `        ``i++; ` `    ``} ` ` `  `    ``// Printing maximum ` `    ``// number of groups ` `    ``cout << no_of_groups << ``"\n"``; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 2, 3, 1, 2, 2 }; ` ` `  `    ``int` `size = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``makeGroups(arr, size); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of above approach ` `import` `java.util.*; ` `class` `GFG{ ` ` `  `// Function that prints the number ` `// of maximum groups ` `static` `void` `makeGroups(``int` `a[], ``int` `n) ` `{ ` `    ``int` `[]v = ``new` `int``[n + ``1``]; ` ` `  `    ``// Store the number of ` `    ``// occurrence of elements ` `    ``for` `(``int` `i = ``0``; i < n; i++) ` `    ``{ ` `        ``v[a[i]]++; ` `    ``} ` ` `  `    ``int` `no_of_groups = ``0``; ` ` `  `    ``// Make all groups of similar ` `    ``// elements and store the ` `    ``// left numbers ` `    ``for` `(``int` `i = ``1``; i <= n; i++)  ` `    ``{ ` `        ``no_of_groups += v[i] / i; ` ` `  `        ``v[i] = v[i] % i; ` `    ``} ` ` `  `    ``int` `i = ``1``; ` `    ``int` `total = ``0``; ` ` `  `    ``for` `(i = ``1``; i <= n; i++)  ` `    ``{ ` `        ``// Condition for finding first ` `        ``// leftover element ` `        ``if` `(v[i] != ``0``) ` `        ``{ ` `            ``total = v[i]; ` `            ``break``; ` `        ``} ` `    ``} ` ` `  `    ``i++; ` ` `  `    ``while` `(i <= n)  ` `    ``{ ` `        ``// Condition for current ` `        ``// leftover element ` `        ``if` `(v[i] != ``0``)  ` `        ``{ ` `            ``total += v[i]; ` ` `  `            ``// Condition if group size ` `            ``// is equal to or more than ` `            ``// current element ` `            ``if` `(total >= i)  ` `            ``{ ` `                ``int` `rem = total - i; ` `                ``no_of_groups++; ` `                ``total = rem; ` `            ``} ` `        ``} ` `        ``i++; ` `    ``} ` ` `  `    ``// Printing maximum ` `    ``// number of groups ` `    ``System.out.print(no_of_groups + ``"\n"``); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `arr[] = { ``2``, ``3``, ``1``, ``2``, ``2` `}; ` ` `  `    ``int` `size = arr.length; ` ` `  `    ``makeGroups(arr, size); ` `} ` `} ` ` `  `// This code is contributed by sapnasingh4991 `

## C#

 `// C# implementation of above approach ` `using` `System; ` `class` `GFG{ ` ` `  `// Function that prints the number ` `// of maximum groups ` `static` `void` `makeGroups(``int` `[]a, ``int` `n) ` `{ ` `    ``int` `[]v = ``new` `int``[n + 1]; ` `    ``int` `i = 0; ` `     `  `    ``// Store the number of ` `    ``// occurrence of elements ` `    ``for``(i = 0; i < n; i++) ` `    ``{ ` `       ``v[a[i]]++; ` `    ``} ` ` `  `    ``int` `no_of_groups = 0; ` ` `  `    ``// Make all groups of similar ` `    ``// elements and store the ` `    ``// left numbers ` `    ``for``(i = 1; i <= n; i++)  ` `    ``{ ` `       ``no_of_groups += v[i] / i; ` `       ``v[i] = v[i] % i; ` `    ``} ` ` `  `    ``i = 1; ` `    ``int` `total = 0; ` `    ``for``(i = 1; i <= n; i++)  ` `    ``{ ` `         `  `       ``// Condition for finding first ` `       ``// leftover element ` `       ``if` `(v[i] != 0) ` `       ``{ ` `           ``total = v[i]; ` `           ``break``; ` `       ``} ` `    ``} ` `    ``i++; ` ` `  `    ``while` `(i <= n)  ` `    ``{ ` `         `  `        ``// Condition for current ` `        ``// leftover element ` `        ``if` `(v[i] != 0)  ` `        ``{ ` `            ``total += v[i]; ` ` `  `            ``// Condition if group size ` `            ``// is equal to or more than ` `            ``// current element ` `            ``if` `(total >= i)  ` `            ``{ ` `                ``int` `rem = total - i; ` `                ``no_of_groups++; ` `                ``total = rem; ` `            ``} ` `        ``} ` `        ``i++; ` `    ``} ` ` `  `    ``// Printing maximum ` `    ``// number of groups ` `    ``Console.Write(no_of_groups + ``"\n"``); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `[]arr = { 2, 3, 1, 2, 2 }; ` `    ``int` `size = arr.Length; ` ` `  `    ``makeGroups(arr, size); ` `} ` `} ` ` `  `// This code is contributed by sapnasingh4991 `

Output:

```2
```

Time Complexity: O(N)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : sapnasingh4991