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 2Input: arr[] = {2, 4, 2, 3}, X = 8
Output: 1
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.
- Sort the array in descending order.
- Initialize variables counter, sz to store the count of possible subsets, and the size of the current subset.
- 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.
- 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 <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 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 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# 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 |
<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>
|
2
Time Complexity: O(NLog(N))
Auxiliary Space: O(1)