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Maximum number of subsets an array can be split into such that product of their minimums with size of subsets is at least K
  • Last Updated : 05 May, 2021

Given an array arr[] consisting of N integers and an integer K, the task is to find the maximum number of disjoint subsets that the given array can be split into such that the product of the minimum element of each subset with the size of the subset is at least K.

Examples:

Input: arr[] = {7, 11, 2, 9, 5}, K = 10
Output: 2
Explanation:
All such disjoint subsets possible are:
Subset {11}: Product of minimum and size of the subset = 11 * 1 = 11 ( > 10).
Subset {5, 9, 7}: Product of minimum and size of the subset = 5 * 3 = 15( > 10).
Therefore, the total number of subsets formed is 2.

Input: arr[] = {1, 3, 3, 7}, K = 12
Output: 0

Approach: The given problem can be solved greedily based on the following observations:



  • As given in the problem statement the product of the minimum element of the formed subset and the length of the subset must be at least K, so to maximize the number of subsets, the maximum element of the array can be grouped to the minimum element of the subset.
  • So the idea is to maximize the minimum element of the subset one by one, which maximizes the count of the subset.

Follow the steps below to solve the problem:

  • Initialize a variable, say count as 0, to store the maximum number of subsets formed.
  • Initialize a variable, say length as 0, to store the length of the subset.
  • Sort the array in descending order.
  • Traverse the given array arr[] and perform the following steps:
    • Increment the value of length by 1.
    • If the value of (arr[i] * length) is greater than K, then increment the value of the count by 1 and update the value of length as 0.
  • After completing the above steps, print the value of count as the resultant maximum number of subsets formed.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the maximum number
// of subsets possible such that
// product of their minimums and the
// size of subsets are at least K
int maximumSubset(int arr[], int N,
                  int K)
{
    // Sort the array in
    // descending order
    sort(arr, arr + N, greater<int>());
 
    // Stores the size of
    // the current subset
    int len = 0;
 
    // Stores the count of subsets
    int ans = 0;
 
    // Traverse the array arr[]
    for (int i = 0; i < N; i++) {
 
        // Increment length of the
        // subsets by 1
        len++;
 
        // If arr[i] * len >= K
        if (arr[i] * len >= K) {
 
            // Increment ans by one
            ans++;
 
            // Update len
            len = 0;
        }
    }
 
    // Return the maximum possible
    // subsets formed
    return ans;
}
 
// Driver Code
int main()
{
    int arr[] = { 7, 11, 2, 9, 5 };
    int K = 10;
    int N = sizeof(arr) / sizeof(arr[0]);
    cout << maximumSubset(arr, N, K);
 
    return 0;
}

Java




import java.util.*;
public class GFG
{
 
  // Function to reverse the sorted array
  public static void reverse(int[] arr)
  {
 
    // Length of the array
    int n = arr.length;
 
    // Swaping the first half elements with last half
    // elements
    for (int i = 0; i < n / 2; i++) {
 
      // Storing the first half elements temporarily
      int temp = arr[i];
 
      // Assigning the first half to the last half
      arr[i] = arr[n - i - 1];
 
      // Assigning the last half to the first half
      arr[n - i - 1] = temp;
    }
  }
 
  // Function to find the maximum number
  // of subsets possible such that
  // product of their minimums and the
  // size of subsets are at least K
  public static int maximumSubset(int arr[], int N, int K)
  {
     
    // Sort the array in
    // descending order
    Arrays.sort(arr);
    reverse(arr);
    // Stores the size of
    // the current subset
    int len = 0;
 
    // Stores the count of subsets
    int ans = 0;
 
    // Traverse the array arr[]
    for (int i = 0; i < N; i++) {
 
      // Increment length of the
      // subsets by 1
      len++;
 
      // If arr[i] * len >= K
      if (arr[i] * len >= K) {
 
        // Increment ans by one
        ans++;
 
        // Update len
        len = 0;
      }
    }
 
    // Return the maximum possible
    // subsets formed
    return ans;
  }
 
  // Driver Code
  public static void main(String args[]) {
    int arr[] = { 7, 11, 2, 9, 5 };
    int K = 10;
    int N =arr.length;
    System.out.println(maximumSubset(arr, N, K));
  }
}
 
// This code is contributed by SoumikMondal

Python3




# Python 3 program for the above approach
 
# Function to find the maximum number
# of subsets possible such that
# product of their minimums and the
# size of subsets are at least K
def maximumSubset(arr, N,
                  K):
 
    # Sort the array in
    # descending order
    arr.sort(reverse = True)
 
    # Stores the size of
    # the current subset
    len = 0
 
    # Stores the count of subsets
    ans = 0
 
    # Traverse the array arr[]
    for i in range(N):
 
        # Increment length of the
        # subsets by 1
        len += 1
 
        # If arr[i] * len >= K
        if (arr[i] * len >= K):
 
            # Increment ans by one
            ans += 1
 
            # Update len
            len = 0
 
    # Return the maximum possible
    # subsets formed
    return ans
 
# Driver Code
if __name__ == "__main__":
 
    arr = [7, 11, 2, 9, 5]
    K = 10
    N = len(arr)
    print(maximumSubset(arr, N, K))
 
    # This code is contributed by ukasp.

C#




using System;
 
public class GFG
{
 
    // Function to reverse the sorted array
    public static void reverse(int[] arr)
    {
 
        // Length of the array
        int n = arr.Length;
 
        // Swaping the first half elements with last half
        // elements
        for (int i = 0; i < n / 2; i++) {
 
            // Storing the first half elements temporarily
            int temp = arr[i];
 
            // Assigning the first half to the last half
            arr[i] = arr[n - i - 1];
 
            // Assigning the last half to the first half
            arr[n - i - 1] = temp;
        }
    }
 
    // Function to find the maximum number
    // of subsets possible such that
    // product of their minimums and the
    // size of subsets are at least K
    public static int maximumSubset(int []arr, int N, int K)
    {
 
        // Sort the array in
        // descending order
        Array.Sort(arr);
        reverse(arr);
       
        // Stores the size of
        // the current subset
        int len = 0;
 
        // Stores the count of subsets
        int ans = 0;
 
        // Traverse the array []arr
        for (int i = 0; i < N; i++)
        {
 
            // Increment length of the
            // subsets by 1
            len++;
 
            // If arr[i] * len >= K
            if (arr[i] * len >= K)
            {
 
                // Increment ans by one
                ans++;
 
                // Update len
                len = 0;
            }
        }
 
        // Return the maximum possible
        // subsets formed
        return ans;
    }
 
    // Driver Code
    public static void Main(String []args)
    {
        int []arr = { 7, 11, 2, 9, 5 };
        int K = 10;
        int N = arr.Length;
        Console.WriteLine(maximumSubset(arr, N, K));
    }
}
 
// This code is contributed by aashish1995.

Javascript




<script>
 
// JavaScript program to implement
// the above approach
 
  // Function to reverse the sorted array
  function reverse(arr)
  {
  
    // Length of the array
    let n = arr.length;
  
    // Swaping the first half elements with last half
    // elements
    for (let i = 0; i < n / 2; i++) {
  
      // Storing the first half elements temporarily
      let temp = arr[i];
  
      // Assigning the first half to the last half
      arr[i] = arr[n - i - 1];
  
      // Assigning the last half to the first half
      arr[n - i - 1] = temp;
    }
  }
  
  // Function to find the maximum number
  // of subsets possible such that
  // product of their minimums and the
  // size of subsets are at least K
  function maximumSubset(arr, N, K)
  {
      
    // Sort the array in
    // descending order
    arr.sort();
    arr.reverse();
    // Stores the size of
    // the current subset
    let len = 0;
  
    // Stores the count of subsets
    let ans = 0;
  
    // Traverse the array arr[]
    for (let i = 0; i < N; i++) {
  
      // Increment length of the
      // subsets by 1
      len++;
  
      // If arr[i] * len >= K
      if (arr[i] * len >= K) {
  
        // Increment ans by one
        ans++;
  
        // Update len
        len = 0;
      }
    }
  
    // Return the maximum possible
    // subsets formed
    return ans;
  }
 
// Driver code
 
    let arr = [ 7, 11, 2, 9, 5 ];
    let K = 10;
    let N =arr.length;
    document.write(maximumSubset(arr, N, K));
            
</script>
Output: 
2

 

Time Complexity: O(N * log N)
Auxiliary Space: O(1)

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