最小化要删除的数组元素的数量，以最大化任意对之间的差异，最大可达K

给定数组arr []和整数K ，任务是计算要从数组中删除的元素的数量，以使剩余的最大数量和最小数量之差不超过K。
例子：

Input: K = 1, arr[] = {1, 2, 3, 4, 5} 
Output: 3 
Explanation: 
Removal of {5, 4, 3} modifies array to {1, 2} where the maximum difference is 1(= K).

Input: K = 3, arr[] = {1, 2, 3, 4, 5} 
Output: 1 
Explanation: 
Removal of {5} modifies array to {1, 2, 3, 4} where the maximum difference is 3( = K).

方法：
任务是找到最大和最小数组元素之间的差，该差不应超过K。

• 以升序对数组进行排序，并将变量初始化为最小值。
• 遍历数组以生成所有可能的对，并检查任何对之间的差是否小于或等于K。
• 更新每对的最小清除次数。
• 最后，打印获得的最小清除量。

下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
#include 
using namespace std;
  
// Function to count the number of
// elements to be removed from the
// array based on the given condition
int min_remove(int arr[], int n, int k)
{
    // Sort the array
    sort(arr, arr + n);
  
    /// Initialize the variable
    int ans = INT_MAX;
  
    // Iterate for all possible pairs
    for (int i = 0; i < n; i++) {
        for (int j = i; j < n; j++) {
  
            // Check the difference
            // between the numbers
            if (arr[j] - arr[i] <= k) {
  
                // Update the minimum removals
                ans = min(ans, n - j + i - 1);
            }
        }
    }
    // Return the answer
    return ans;
}
  
// Driver Code
int main()
{
    int k = 3;
    int arr[] = { 1, 2, 3, 4, 5 };
  
    int n = sizeof arr / sizeof arr[0];
  
    cout << min_remove(arr, n, k);
    return 0;
}

Java

// Java Program to implement
// the above approach
import java.util.*;
class GFG{
  
// Function to count the number of
// elements to be removed from the
// array based on the given condition
static int min_remove(int arr[], int n, int k)
{
    // Sort the array
    Arrays.sort(arr);
  
    /// Initialize the variable
    int ans = Integer.MAX_VALUE;
  
    // Iterate for all possible pairs
    for (int i = 0; i < n; i++) 
    {
        for (int j = i; j < n; j++)
        {
  
            // Check the difference
            // between the numbers
            if (arr[j] - arr[i] <= k) 
            {
  
                // Update the minimum removals
                ans = Math.min(ans, n - j + i - 1);
            }
        }
    }
    // Return the answer
    return ans;
}
  
// Driver Code
public static void main(String[] args)
{
    int k = 3;
    int arr[] = { 1, 2, 3, 4, 5 };
  
    int n = arr.length;
  
    System.out.print(min_remove(arr, n, k));
}
}
  
// This code is contributed by sapnasingh4991

Python3

# Python3 program to implement
# the above approach
import sys
  
# Function to count the number of
# elements to be removed from the
# array based on the given condition
def min_remove(arr, n, k):
  
    # Sort the array
    arr.sort()
  
    # Initialize the variable
    ans = sys.maxsize
  
    # Iterate for all possible pairs
    for i in range(n):
        for j in range(i, n):
  
            # Check the difference
            # between the numbers
            if (arr[j] - arr[i] <= k):
  
                # Update the minimum removals
                ans = min(ans, n - j + i - 1)
      
    # Return the answer
    return ans
  
# Driver Code
if __name__ == "__main__":
  
    k = 3
    arr = [ 1, 2, 3, 4, 5 ]
  
    n = len(arr)
  
    print (min_remove(arr, n, k))
  
# This code is contributed by chitranayal

C#

// C# Program to implement
// the above approach
using System;
class GFG{
  
// Function to count the number of
// elements to be removed from the
// array based on the given condition
static int min_remove(int []arr, int n, int k)
{
    // Sort the array
    Array.Sort(arr);
  
    /// Initialize the variable
    int ans = int.MaxValue;
  
    // Iterate for all possible pairs
    for (int i = 0; i < n; i++) 
    {
        for (int j = i; j < n; j++)
        {
  
            // Check the difference
            // between the numbers
            if (arr[j] - arr[i] <= k) 
            {
  
                // Update the minimum removals
                ans = Math.Min(ans, n - j + i - 1);
            }
        }
    }
    // Return the answer
    return ans;
}
  
// Driver Code
public static void Main(String[] args)
{
    int k = 3;
    int []arr = { 1, 2, 3, 4, 5 };
  
    int n = arr.Length;
  
    Console.Write(min_remove(arr, n, k));
}
}
  
// This code is contributed by sapnasingh4991

输出：

1

时间复杂度： O(N ² )
辅助空间： O(1)