最大化形成的组的数量，其大小不小于其最大元素

给定N个正整数(1≤arr [i]≤N)的数组arr [] ，请将数组的元素划分为组，以使每个组的大小大于或等于该组的最大元素。元素也可能无法加入任何组。任务是最大化组的数量。
例子：

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 
 

方法：

• 首先，将每个元素的出现次数存储在数组中。
• 现在，将相似元素组成组。例如：如果数组中有三个1，则每个1组成三个组。
• 然后存储其余元素，并从最低元素开始分组。

下面是上述方法的实现。

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 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

输出：

2

时间复杂度： O(N)