选择k个数组元素，以最大和最小的差异最小化

给定一个由n个整数和一个正数k组成的数组。我们被允许从给定的数组中获取任何k个整数。任务是找到K个数的最大值和最小值之间的差的最小可能值。

例子：

Input : arr[] = {10, 100, 300, 200, 1000, 20, 30}
        k = 3
Output : 20
20 is the minimum possible difference between any
maximum and minimum of any k numbers.
Given k = 3, we get the result 20 by selecting 
integers {10, 20, 30}.
max(10, 20, 30) - max(10, 20, 30) = 30 - 10 = 20.

Input : arr[] = {1, 2, 3, 4, 10, 20, 30, 40, 
                                   100, 200}.
        k = 4      
Output : 3

这个想法是对数组排序并选择k个连续整数。为什么要连续?令所选的k个整数为arr [0]，arr [1]，.. arr [r]，arr [r + x]…，arr [k-1]，它们在排序数组中的顺序都是递增的，但不连续。这意味着在arr [r]和arr [r + x]之间存在一个整数p。因此，如果包含p并删除arr [0]，则新的差异将是arr [r] – arr [1]，而旧的差异将是arr [r] – arr [0]。而且我们知道arr [0] <= arr [1] <= .. <= arr [k-1]，因此最小差异减小或保持不变。如果我们对其他p个数字执行相同的过程，则得到的差异最小。

解决问题的算法：

1. 对数组进行排序。
2. 计算每组k个连续整数的最大值(k个数)-最小值(k个数)。
3. 返回在步骤2中获得的所有值的最小值。

下面是上述想法的实现：

C++

// C++ program to find minimum difference of maximum
// and minimum of K number.
#include
using namespace std;
  
// Return minimum difference of maximum and minimum
// of k elements of arr[0..n-1].
int minDiff(int arr[], int n, int k)
{
    int result = INT_MAX;
  
    // Sorting the array.
    sort(arr, arr + n);
  
    // Find minimum value among all K size subarray.
    for (int i=0; i<=n-k; i++)
        result = min(result, arr[i+k-1] - arr[i]);
  
    return result;
}
  
// Driven Program
int main()
{
    int arr[] = {10, 100, 300, 200, 1000, 20, 30};
    int n = sizeof(arr)/sizeof(arr[0]);
    int k = 3;
  
    cout << minDiff(arr, n, k) << endl;
    return 0;
}

Java

// Java program to find minimum difference 
// of maximum and minimum of K number.
import java.util.*;
  
class GFG {
      
// Return minimum difference of 
// maximum and minimum of k 
// elements of arr[0..n-1].
static int minDiff(int arr[], int n, int k) {
    int result = Integer.MAX_VALUE;
  
    // Sorting the array.
    Arrays.sort(arr);
  
    // Find minimum value among 
    // all K size subarray.
    for (int i = 0; i <= n - k; i++)
    result = Math.min(result, arr[i + k - 1] - arr[i]);
  
    return result;
}
  
// Driver code
public static void main(String[] args) {
    int arr[] = {10, 100, 300, 200, 1000, 20, 30};
    int n = arr.length;
    int k = 3;
  
    System.out.println(minDiff(arr, n, k));
}
}
  
// This code is contributed by Anant Agarwal.

Python3

# Python program to find minimum
# difference of maximum
# and minimum of K number.
  
# Return minimum difference
# of maximum and minimum
# of k elements of arr[0..n-1].
def minDiff(arr,n,k):
    result = [phone]
   
    # Sorting the array.
    arr.sort()
   
    # Find minimum value among
    # all K size subarray.
    for i in range(n-k+1):
        result = int(min(result, arr[i+k-1] - arr[i]))
   
    return result
  
# Driver code
  
arr= [10, 100, 300, 200, 1000, 20, 30]
n =len(arr)
k = 3
   
print(minDiff(arr, n, k))
  
# This code is contributed
# by Anant Agarwal.

C#

// C# program to find minimum
// difference of maximum and 
// minimum of K number.
using System;
  
class GFG {
      
// Return minimum difference of 
// maximum and minimum of k 
// elements of arr[0..n - 1].
static int minDiff(int []arr, int n,
                   int k) 
{
    int result = int.MaxValue;
  
    // Sorting the array.
    Array.Sort(arr);
  
    // Find minimum value among 
    // all K size subarray.
    for (int i = 0; i <= n - k; i++)
    result = Math.Min(result, arr[i + k - 1] - arr[i]);
  
    return result;
}
  
// Driver code
public static void Main() {
    int []arr = {10, 100, 300, 200, 1000, 20, 30};
    int n = arr.Length;
    int k = 3;
  
    Console.WriteLine(minDiff(arr, n, k));
}
}
  
// This code is contributed by vt_m.

输出：

20

时间复杂度： O(nlogn)。