Remove minimum elements from the array such that 2*min becomes more than max

Given an array of size N. The task is to remove minimum elements from the array such that twice of minimum number is greater than the maximum number in the modified array. Print the minimum number of elements removed.

Examples:

Input: arr[] = {4, 5, 100, 9, 10, 11, 12, 15, 200}
Output: 4
Remove 4 elements (4, 5, 100, 200)
so that 2*min becomes more than max.

Input: arr[] = {4, 7, 5, 6}
Output: 0

Approach:

  • sort the given array
  • Traverse from left to right in the array and for each element chosen (let it be x) with the index i, find the upper_bound of (2*x). let that index be j.Then, update the our required answer by (n-j+i) if (n-j+i) is less than current value of our answer .

Below is the implementation of the above approach :

C++

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// CPP program to remove minimum elements from the
// array such that 2*min becomes more than max
#include <bits/stdc++.h>
using namespace std;
  
// Function to remove minimum elements from the
// array such that 2*min becomes more than max
int Removal(vector<int> v, int n)
{
    // Sort the array
    sort(v.begin(), v.end());
  
    // To store the required answer
    int ans = INT_MAX;
  
    // Traverse from left to right
    for (vector<int>::iterator i = v.begin(); i != v.end(); 
                                                     i++) {
  
        vector<int>::iterator j = upper_bound(v.begin(), 
                                    v.end(), (2 * (*i)));
  
        // Update the answer
        ans = min(ans, n - (int)(j - i));
    }
  
    // Return the required answer
    return ans;
}
  
// Driver code
int main()
{
    vector<int> a = { 4, 5, 100, 9, 10, 11, 12, 15, 200 };
  
    int n = a.size();
  
    // Function call
    cout << Removal(a, n);
  
    return 0;
}

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Java

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// Java program to remove minimum elements from the 
// array such that 2*min becomes more than max
import java.util.Arrays;
  
class GFG 
{
  
    // Function to calculate upper bound
    public static int upperBound(int[] array,
                                 int value)
    {
        int low = 0;
        int high = array.length;
        while (low < high) 
        {
            final int mid = (low + high) / 2;
            if (value >= array[mid]) 
            {
                low = mid + 1;
            
            else 
            {
                high = mid;
            }
        }
        return low;
    }
  
    // Function to remove minimum elements from the
    // array such that 2*min becomes more than max
    public static int Removal(int[] v, int n) 
    {
  
        // Sort the array
        Arrays.sort(v);
  
        // To store the required answer
        int ans = Integer.MAX_VALUE;
        int k = 0;
  
        // Traverse from left to right
        for (int i : v) 
        {
            int j = upperBound(v, (2 * i));
  
            // Update the answer
            ans = Math.min(ans, n - (j - k));
            k++;
        }
  
        // Return the required answer
        return ans;
    }
  
    // Driver code
    public static void main(String[] args) 
    {
        int[] a = { 4, 5, 100, 9, 10
                    11, 12, 15, 200 };
        int n = a.length;
  
        // Function call
        System.out.println(Removal(a, n));
    }
}
  
// This code is contributed by
// sanjeev2552

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C#

// C# program to remove minimum elements
// from the array such that 2*min becomes
// more than max
using System;

class GFG
{

// Function to calculate upper bound
public static int upperBound(int[] array,
int value)
{
int low = 0;
int high = array.Length;
while (low < high) { int mid = (low + high) / 2; if (value >= array[mid])
{
low = mid + 1;
}
else
{
high = mid;
}
}
return low;
}

// Function to remove minimum elements from the
// array such that 2*min becomes more than max
public static int Removal(int[] v, int n)
{

// Sort the array
Array.Sort(v);

// To store the required answer
int ans = int.MaxValue;
int k = 0;

// Traverse from left to right
foreach (int i in v)
{
int j = upperBound(v, (2 * i));

// Update the answer
ans = Math.Min(ans, n – (j – k));
k++;
}

// Return the required answer
return ans;
}

// Driver code
public static void Main(String[] args)
{
int[] a = { 4, 5, 100, 9, 10,
11, 12, 15, 200 };
int n = a.Length;

// Function call
Console.WriteLine(Removal(a, n));
}
}

// This code is contributed by Rajput-Ji

Output :

4

Time complexity: O(NlogN)



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Improved By : sanjeev2552, Rajput-Ji