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

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

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++ Program to implement ` `// the above approach ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Comperator 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 ` `// satisy 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 program to implement ` `// the above approach ` `import` `java.util.*; ` ` `  `class` `GFG{ ` ` `  `// Function to return the maximum count ` `// of subsets possible which ` `// satisy 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 Program to implement ` `# the above approach ` ` `  `# Function to return the maximum count ` `# of subsets possible which ` `# satisy 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# program to implement ` `// the above approach ` `using` `System; ` `using` `System.Linq; ` ` `  `class` `GFG{ ` ` `  `// Function to return the maximum count ` `// of subsets possible which ` `// satisy 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`

Output:
```2
```

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

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