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Maximise count of elements that are strictly greater in a subsequence than their average

  • Last Updated : 16 Dec, 2021

Given an array arr[] of size N containing positive integers, the task is to find the maximum number of elements that can be deleted from the array using any number of operations. In one operation, select a subsequence from the given array, take their average and delete the numbers which are strictly greater than that average from the array.

Example:

Input: arr[] = {1, 1, 3, 2, 4}
Output: 3
Explanation:
Operation 1: Choose the subsequence {1, 2, 4}, average = (1+2+4)/3 = 2. So arr[5]=4 is deleted. arr[]={1, 1, 3, 2}
Operation 2: Choose the subsequence {1, 3, 2}, average = (1+3+2)/3 = 2. So arr[2]=3 is deleted. arr[]={1, 1, 2}
Operation 3: Choose the subsequence {1, 1}, average = (1+1)/2 = 1. So arr[3]=2 is deleted. arr[]={1, 1}
No further deletions can be performed.

Input: arr[] = {5, 5, 5}
Output: 0

Approach: The catch in this problem is that all elements except the minimum one can be deleted from the array because if only the minimum element is used to create the subsequence, then its average is basically the same element and all the other elements can be deleted. Now to solve this problem, follow the below steps:

  1. Find the frequency of minimum element, say freq.
  2. Return N-freq as the answer to this problem.

Below is the implementation of the above approach:

C++




// C++ code for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the maximum number of
// elements that can be deleted from the array
int elementsDeleted(vector<int>& arr)
{
 
    // Size of the array
    int N = arr.size();
 
    // Minimum element
    auto it = *min_element(arr.begin(), arr.end());
 
    // Finding frequency of minimum element
    int freq = 0;
    for (auto x : arr) {
        if (x == it)
            freq++;
    }
 
    return N - freq;
}
 
// Driver Code
int main()
{
    vector<int> arr = { 3, 1, 1, 2, 4 };
    cout << elementsDeleted(arr);
}


Java




// Java code for the above approach
import java.util.*;
 
class GFG{
 
// Function to find the maximum number of
// elements that can be deleted from the array
static int elementsDeleted(int []arr)
{
 
    // Size of the array
    int N = arr.length;
 
    // Minimum element
    int it = Arrays.stream(arr).min().getAsInt();
 
    // Finding frequency of minimum element
    int freq = 0;
    for (int x : arr) {
        if (x == it)
            freq++;
    }
 
    return N - freq;
}
 
// Driver Code
public static void main(String[] args)
{
    int []arr = { 3, 1, 1, 2, 4 };
    System.out.print(elementsDeleted(arr));
}
}
 
// This code is contributed by 29AjayKumar


Python3




# Python code for the above approach
 
# Function to find the maximum number of
# elements that can be deleted from the array
def elementsDeleted(arr):
 
    # Size of the array
    N = len(arr)
 
    # Minimum element
    it = 10**9
 
    for i in range(len(arr)):
        it = min(it, arr[i])
 
    # Finding frequency of minimum element
    freq = 0
    for x in arr:
        if (x == it):
            freq += 1
    return N - freq
 
# Driver Code
arr = [3, 1, 1, 2, 4]
print(elementsDeleted(arr))
 
# This code is contributed by Saurabh jaiswal


C#




// C# code for the above approach
using System;
using System.Linq;
class GFG {
 
    // Function to find the maximum number of
    // elements that can be deleted from the array
    static int elementsDeleted(int[] arr)
    {
 
        // Size of the array
        int N = arr.Length;
 
        // Minimum element
        int it = arr.Min();
 
        // Finding frequency of minimum element
        int freq = 0;
        foreach(int x in arr)
        {
            if (x == it)
                freq++;
        }
 
        return N - freq;
    }
 
    // Driver Code
    public static void Main(string[] args)
    {
        int[] arr = { 3, 1, 1, 2, 4 };
        Console.WriteLine(elementsDeleted(arr));
    }
}
 
// This code is contributed by ukasp.


Javascript




<script>
      // JavaScript code for the above approach
 
      // Function to find the maximum number of
      // elements that can be deleted from the array
      function elementsDeleted(arr) {
 
          // Size of the array
          let N = arr.length;
 
          // Minimum element
          let it = Number.MAX_VALUE;
 
          for (let i = 0; i < arr.length; i++) {
              it = Math.min(it, arr[i]);
          }
 
          // Finding frequency of minimum element
          let freq = 0;
          for (let x of arr) {
              if (x == it)
                  freq++;
          }
 
          return N - freq;
      }
 
      // Driver Code
 
      let arr = [3, 1, 1, 2, 4];
      document.write(elementsDeleted(arr));
 
 
// This code is contributed by Potta Lokesh
  </script>


 
 

Output

3

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


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