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Maximize count of subsets having product of smallest element and size of the subset at least X

  • Last Updated : 04 Sep, 2021

Given an array arr[] consisting of N integers and an integer X, the task is to count the maximum number of subsets possible from the given array having  

Smallest element of the Subset * Size of the Subset ≥ X

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 Examples: 



Input: arr[] = {7, 11, 2, 9, 5}, X = 10
Output: 2
Explanation: 
One of the possible solution is {7, 9} and {11, 5}. 
In subset {7, 9} the smallest element (= 7), size of subset (= 2). 
Therefore, the product of the smallest element of the subset and the size of the subset is equal to 14 ( > 10). 
In subset {11, 5} the smallest element (= 5), size of subset (= 2). 
Therefore, the product of the smallest element of the subset * size of the subset is equal to 10 
Hence, the required output is 2

Input: arr[] = {2, 4, 2, 3}, X = 8
Output:

Approach: The problem can be solved using Greedy Approach. The idea is to sort the array in descending order and then traverse the elements one by one to check the required conditions.
Follow the below steps to solve the problem. 

  1. Sort the array in descending order.
  2. Initialize variables counter, sz to store the count of possible subsets, and the size of the current subset.
  3. Iterate over the given array and check if arr[i] * sz ≥ X or not. If found to be true, reset the sz and increment counter by 1.
  4. Finally, after complete traversal of the array, print counter as the required answer.

Below is the implementation of the above approach:

C++




// C++ Program to implement
// the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Comparator function to return
// the greater of two numbers
bool comp(int a, int b)
{
    return a > b;
}
 
// Function to return the maximum count
// of subsets possible which
// satisfy the above condition
int maxSubset(int arr[], int N, int X)
{
 
    // Sort the array in
    // descending order
    sort(arr, arr + N, comp);
 
    // Stores the count of subsets
    int counter = 0;
 
    // Stores the size of
    // the current subset
    int sz = 0;
 
    for (int i = 0; i < N; i++)
    {
        sz++;
 
        // Check for the necessary
        // conditions
        if (arr[i] * sz >= X) {
            counter++;
            sz = 0;
        }
    }
    return counter;
}
 
// Driver Code
int main()
{
    int arr[] = { 7, 11, 2, 9, 5 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int X = 10;
    cout << maxSubset(arr, N, X);
}

Java




// Java program to implement
// the above approach
import java.util.*;
 
class GFG{
 
// Function to return the maximum count
// of subsets possible which
// satisfy the above condition
static int maxSubset(Integer arr[], int N,
                                    int X)
{
     
    // Sort the array in
    // descending order
    Arrays.sort(arr, Collections.reverseOrder());
 
    // Stores the count of subsets
    int counter = 0;
 
    // Stores the size of
    // the current subset
    int sz = 0;
 
    for(int i = 0; i < N; i++)
    {
        sz++;
 
        // Check for the necessary
        // conditions
        if (arr[i] * sz >= X)
        {
            counter++;
            sz = 0;
        }
    }
    return counter;
}
 
// Driver Code
public static void main(String[] args)
{
    Integer arr[] = { 7, 11, 2, 9, 5 };
    int N = arr.length;
    int X = 10;
     
    System.out.print(maxSubset(arr, N, X));
}
}
 
// This code is contributed by 29AjayKumar

Python3




# Python3 Program to implement
# the above approach
 
# Function to return the maximum count
# of subsets possible which
# satisfy the above condition
def maxSubset(arr, N, X):
 
    # Sort the array in
    # descending order
    arr.sort(reverse = True)
 
    # Stores the count of subsets
    counter = 0
 
    # Stores the size of
    # the current subset
    sz = 0
 
    for i in range(N):
        sz += 1
 
        # Check for the necessary
        # conditions
        if(arr[i] * sz >= X):
            counter += 1
            sz = 0
 
    return counter
 
# Driver Code
 
# Given array
arr = [ 7, 11, 2, 9, 5 ]
N = len(arr)
X = 10
 
# Function call
print(maxSubset(arr, N, X))
 
# This code is contributed by Shivam Singh

C#




// C# program to implement
// the above approach
using System;
using System.Linq;
 
class GFG{
 
// Function to return the maximum count
// of subsets possible which
// satisfy the above condition
static int maxSubset(int []arr, int N,
                                int X)
{
     
    // Sort the array in
    // descending order
    Array.Sort(arr);
    Array.Reverse(arr);
 
    // Stores the count of subsets
    int counter = 0;
 
    // Stores the size of
    // the current subset
    int sz = 0;
 
    for(int i = 0; i < N; i++)
    {
        sz++;
 
        // Check for the necessary
        // conditions
        if (arr[i] * sz >= X)
        {
            counter++;
            sz = 0;
        }
    }
    return counter;
}
 
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 7, 11, 2, 9, 5 };
    int N = arr.Length;
    int X = 10;
     
    Console.Write(maxSubset(arr, N, X));
}
}
 
// This code is contributed by shikhasingrajput

Javascript




<script>
// Javascript Program to implement
// the above approach
 
// Comparator function to return
// the greater of two numbers
function comp(a, b)
{
    return a > b;
}
 
// Function to return the maximum count
// of subsets possible which
// satisfy the above condition
function maxSubset(arr, N, X)
{
 
    // Sort the array in
    // descending order
    arr = arr.sort((a, b)=>b-a);
 
    // Stores the count of subsets
    let counter = 0;
 
    // Stores the size of
    // the current subset
    let sz = 0;
 
    for (let i = 0; i < N; i++)
    {
        sz++;
 
        // Check for the necessary
        // conditions
        if (arr[i] * sz >= X) {
            counter++;
            sz = 0;
        }
    }
    return counter;
}
 
// Driver Code
    let arr = [ 7, 11, 2, 9, 5 ];
    let N = arr.length;
    let X = 10;
    document.write(maxSubset(arr, N, X));
 
// This code is contributed by Saurabh Jaiswal
    </script>
Output: 
2

Time Complexity: O(NLog(N))
Auxiliary Space: O(1)




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