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 

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 <bits/stdc++.h>
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

<script>

// Javascript implementation of above approach

// Function that prints the number
// of maximum groups
function makeGroups(a, n)
{
    let v = Array.from({length: n+1}, (_, i) => 0);

    // Store the number of
    // occurrence of elements
    for (let i = 0; i < n; i++)
    {
        v[a[i]]++;
    }

    let no_of_groups = 0;

    // Make all groups of similar
    // elements and store the
    // left numbers
    for (let i = 1; i <= n; i++)
    {
        no_of_groups += Math.floor(v[i] / i);

        v[i] = v[i] % i;
    }

    let i = 1;
    let 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)
            {
                let rem = total - i;
                no_of_groups++;
                total = rem;
            }
        }
        i++;
    }

    // Printing maximum
    // number of groups
    document.write(no_of_groups + "\n");
}

// Driver Code

   let arr = [ 2, 3, 1, 2, 2 ];

    let size = arr.length;

    makeGroups(arr, size);

</script>

Output:

2

Time Complexity: O(N) 

Auxiliary Space: O(N)