将给定数组拆分为K个子数组，以最大和最小值之间的差异最小化

给定N个整数和整数K的排序数组arr [] ，任务是将数组拆分为K个子数组，以使每个子数组的最大元素和最小元素之差之和最小。

例子：

Input: arr[] = {1, 3, 3, 7}, K = 4
Output: 0
Explanation:
The given array can be split into 4 subarrays as {1}, {3}, {3}, and {7}.
The difference between minimum and maximum of each subarray is:
1. {1}, difference = 1 – 1 = 0
2. {3}, difference = 3 – 3 = 0
3. {3}, difference = 3 – 3 = 0
4. {7}, difference = 7 – 7 = 0
Therefore, the sum all the difference is 0 which is minimized.

Input: arr[] = {4, 8, 15, 16, 23, 42}, K = 3
Output: 12
Explanation:
The given array can be split into 3 subarrays as {4, 8, 15, 16}, {23}, and {42}.
The difference between minimum and maximum of each subarray is:
1. {4, 8, 15, 16}, difference = 16 – 12 = 0
2. {23}, difference = 23 – 23 = 0
3. {42}, difference = 42 – 42 = 0
Therefore, the sum all the difference is 12 which is minimized.

方法：在给定条件下将给定数组拆分为K个子数组，其思想是在元素arr [i + 1]和arr [i]之间的差异最大的索引处(例如i )进行拆分。以下是实现此方法的步骤：

1. 将给定数组arr []中连续的元素对之间的差异存储到另一个数组中(例如temp [] )。
2. 对数组temp []进行升序排序。
3. 将总差(例如diff )初始化为给定数组arr []的第一个元素和最后一个元素的差。
4. 将数组temp []中的前K – 1个值添加到上述差异中。
5. diff中存储的值是K子数组的最大元素和最小元素之差的最小和。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
  
// Function to find the subarray
int find(int a[], int n, int k)
{
    vector v;
  
    // Add the difference to vectors
    for (int i = 1; i < n; ++i) {
        v.push_back(a[i - 1] - a[i]);
    }
  
    // Sort vector to find minimum k
    sort(v.begin(), v.end());
  
    // Initialize result
    int res = a[n - 1] - a[0];
  
    // Adding first k-1 values
    for (int i = 0; i < k - 1; ++i) {
        res += v[i];
    }
  
// Return the minimized sum
    return res;
}
  
// Driver Code
int main()
{
// Given array arr[]
    int arr[] = { 4, 8, 15, 16, 23, 42 };
  
    int N = sizeof(arr) / sizeof(int);
  
// Given K
int K = 3;
  
// Function Call
    cout << find(arr, N, K) << endl;
  
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
  
class GFG{
  
// Function to find the subarray
static int find(int a[], int n, int k)
{
    Vector v = new Vector();
  
    // Add the difference to vectors
    for(int i = 1; i < n; ++i)
    {
       v.add(a[i - 1] - a[i]);
    }
  
    // Sort vector to find minimum k
    Collections.sort(v);
  
    // Initialize result
    int res = a[n - 1] - a[0];
  
    // Adding first k-1 values
    for(int i = 0; i < k - 1; ++i)
    {
       res += v.get(i);
    }
      
    // Return the minimized sum
    return res;
}
  
// Driver Code
public static void main(String[] args)
{
      
    // Given array arr[]
    int arr[] = { 4, 8, 15, 16, 23, 42 };
  
    int N = arr.length;
      
    // Given K
    int K = 3;
      
    // Function Call
    System.out.print(find(arr, N, K) + "\n");
}
}
  
// This code is contributed by Amit Katiyar

Python3

# Python3 program for the above approach
  
# Function to find the subarray
def find(a, n, k):
      
    v = []
      
    # Add the difference to vectors
    for i in range(1, n):
        v.append(a[i - 1] - a[i])
          
    # Sort vector to find minimum k
    v.sort()
      
    # Initialize result
    res = a[n - 1] - a[0]
      
    # Adding first k-1 values
    for i in range(k - 1):
        res += v[i]
      
    # Return the minimized sum
    return res
      
# Driver code
arr = [ 4, 8, 15, 16, 23, 42 ]
  
# Length of array
N = len(arr)
  
K = 3
  
# Function Call
print(find(arr, N, K))
  
# This code is contributed by sanjoy_62

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
  
class GFG{
  
// Function to find the subarray
static int find(int []a, int n, int k)
{
    List v = new List();
  
    // Add the difference to vectors
    for(int i = 1; i < n; ++i)
    {
       v.Add(a[i - 1] - a[i]);
    }
  
    // Sort vector to find minimum k
    v.Sort();
  
    // Initialize result
    int res = a[n - 1] - a[0];
  
    // Adding first k-1 values
    for(int i = 0; i < k - 1; ++i)
    {
       res += v[i];
    }
      
    // Return the minimized sum
    return res;
}
  
// Driver Code
public static void Main(String[] args)
{
      
    // Given array []arr
    int []arr = { 4, 8, 15, 16, 23, 42 };
    int N = arr.Length;
      
    // Given K
    int K = 3;
      
    // Function Call
    Console.Write(find(arr, N, K) + "\n");
}
}
  
// This code is contributed by Amit Katiyar

输出：

12

时间复杂度： O(N) ，其中N是数组中元素的数量。
辅助空间： O(N) ，其中N是数组中元素的数量。