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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:
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:

Approach:

• Firstly store the number of occurrences 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 the 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[0]);` `    ``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`

Python3

 `# python3 implementation of above approach``# Function that prints the number``# of maximum groups``def` `makeGroups(a, n):``    ``v ``=` `[``0``] ``*` `(n ``+` `1``)` `    ``# Store the number of``    ``# occurrence of elements``    ``for` `i ``in` `range` `(n):``        ``v[a[i]] ``+``=` `1``   ` `    ``no_of_groups ``=` `0` `    ``# Make all groups of similar``    ``# elements and store the``    ``# left numbers``    ``for` `i ``in` `range` `(``1``, n ``+` `1``):``        ``no_of_groups ``+``=` `v[i] ``/``/` `i``        ``v[i] ``=` `v[i] ``%` `i``   ` `    ``i ``=` `1``    ``total ``=` `0``    ``for` `i ``in` `range` `( ``1``, n ``+` `1``):``      ` `        ``# Condition for finding first``        ``# leftover element``        ``if` `(v[i] !``=` `0``):``            ``total ``=` `v[i]``            ``break` `    ``i ``+``=` `1``    ``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):``                ``rem ``=` `total ``-` `i``                ``no_of_groups ``+``=` `1``                ``total ``=` `rem``        ` `        ``i ``+``=` `1` `    ``# Printing maximum``    ``# number of groups``    ``print` `(no_of_groups)` `# Driver Code``if` `__name__ ``=``=` `"__main__"``: ``    ``arr ``=` `[``2``, ``3``, ``1``, ``2``, ``2``]``    ``size ``=` `len``(arr)``    ``makeGroups(arr, size)` `# This code is contributed by Chitranayal`

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`

Javascript

 ``

Output:

`2`

Time Complexity: O(N)

Auxiliary Space: O(N)

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