给定大小为N且整数K的数组arr [] ，任务是将数组拆分为K个子数组，以最小化每个子数组的相邻元素之间的绝对差之和。
例子：
Input: arr[] = {1, 3, -2, 5, -1}, K = 2
Output: 13
Explanation: Split the array into following 2 subarrays: {1, 3, -2} and {5, -1}.
Input: arr[] = {2, 14, 26, 10, 5, 12}, K = 3
Output: 24
Explanation: Splitting array into following 3 subarrays: {2, 14}, {26}, {10, 5, 12}.
方法：可以根据以下观察结果解决给定问题：
The idea is to slice the array arr[] at ith index which gives maximum absolute difference of adjacent elements. Subtract it from the result. Slicing at K – 1 places will give K subarrays with minimum sum of absolute difference of adjacent elements.
请按照以下步骤解决问题：
- 初始化一个数组，例如new_Arr []和一个整数变量，例如ans ，以存储总的绝对差和。
- 遍历数组。
- 将相邻元素的绝对差(例如arr [i + 1]和arr [i])存储在new_Arr []数组中。
- 通过arr [i]和arr [i + 1]的绝对差值增加ans
- 按降序对new_Arr []数组进行排序。
- 从i = 0遍历到i = K-1
- 用new_Arr [i]减少ans 。
- 最后，打印出ans。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to split an array into K subarrays
// with minimum sum of absolute difference
// of adjacent elements in each of K subarrays
void absoluteDifference(int arr[], int N, int K)
{
// Stores the absolute differences
// of adjacent elements
int new_Arr[N - 1];
// Stores the answer
int ans = 0;
// Stores absolute differences of
// adjacent elements in new_Arr
for (int i = 0; i < N - 1; i++) {
new_Arr[i] = abs(arr[i] - arr[i + 1]);
// Stores the sum of all absolute
// differences of adjacent elements
ans += new_Arr[i];
}
// Sorting the new_Arr
// in decreasing order
sort(new_Arr, new_Arr + N - 1,
greater());
for (int i = 0; i < K - 1; i++) {
// Removing K - 1 elements
// with maximum sum
ans -= new_Arr[i];
}
// Prints the answer
cout << ans << endl;
}
// Driver code
int main()
{
// Given array arr[]
int arr[] = { 1, 3, -2, 5, -1 };
// Given K
int K = 2;
// Size of array
int N = sizeof(arr) / sizeof(arr[0]);
// Function Call
absoluteDifference(arr, N, K);
return 0;
} Java
// java program for the above approach
import java.util.*;
import java.util.Arrays;
class GFG
{
public static void reverse(int[] array)
{
// Length of the array
int n = array.length;
// Swaping the first half elements with last half
// elements
for (int i = 0; i < n / 2; i++)
{
// Storing the first half elements temporarily
int temp = array[i];
// Assigning the first half to the last half
array[i] = array[n - i - 1];
// Assigning the last half to the first half
array[n - i - 1] = temp;
}
}
// Function to split an array into K subarrays
// with minimum sum of absolute difference
// of adjacent elements in each of K subarrays
public static void absoluteDifference(int arr[],
int N, int K)
{
// Stores the absolute differences
// of adjacent elements
int new_Arr[] = new int[N - 1];
// Stores the answer
int ans = 0;
// Stores absolute differences of
// adjacent elements in new_Arr
for (int i = 0; i < N - 1; i++)
{
new_Arr[i] = Math.abs(arr[i] - arr[i + 1]);
// Stores the sum of all absolute
// differences of adjacent elements
ans += new_Arr[i];
}
// Sorting the new_Arr
// in decreasing order
Arrays.sort(new_Arr);
reverse(new_Arr);
for (int i = 0; i < K - 1; i++)
{
// Removing K - 1 elements
// with maximum sum
ans -= new_Arr[i];
}
// Prints the answer
System.out.println(ans);
}
// Driver code
public static void main (String[] args)
{
// Given array arr[]
int arr[] = { 1, 3, -2, 5, -1 };
// Given K
int K = 2;
// Size of array
int N = arr.length;
// Function Call
absoluteDifference(arr, N, K);
}
}
// This code is contributed by AnkThonPython3
# Python3 program for the above approach
# Function to split an array into K subarrays
# with minimum sum of absolute difference
# of adjacent elements in each of K subarrays
def absoluteDifference(arr, N, K):
# Stores the absolute differences
# of adjacent elements
new_Arr = [0 for i in range(N - 1)]
# Stores the answer
ans = 0
# Stores absolute differences of
# adjacent elements in new_Arr
for i in range(N - 1):
new_Arr[i] = abs(arr[i] - arr[i + 1])
# Stores the sum of all absolute
# differences of adjacent elements
ans += new_Arr[i]
# Sorting the new_Arr
# in decreasing order
new_Arr = sorted(new_Arr)[::-1]
for i in range(K - 1):
# Removing K - 1 elements
# with maximum sum
ans -= new_Arr[i]
# Prints the answer
print (ans)
# Driver code
if __name__ == '__main__':
# Given array arr[]
arr = [ 1, 3, -2, 5, -1 ]
# Given K
K = 2
# Size of array
N = len(arr)
# Function Call
absoluteDifference(arr, N, K)
# This code is contributed by mohit kumar 29C#
// C# program for the above approach
using System;
class GFG{
public static void reverse(int[] array)
{
// Length of the array
int n = array.Length;
// Swaping the first half elements with
// last half elements
for(int i = 0; i < n / 2; i++)
{
// Storing the first half elements
// temporarily
int temp = array[i];
// Assigning the first half to
// the last half
array[i] = array[n - i - 1];
// Assigning the last half to
// the first half
array[n - i - 1] = temp;
}
}
// Function to split an array into K subarrays
// with minimum sum of absolute difference
// of adjacent elements in each of K subarrays
public static void absoluteDifference(int[] arr,
int N, int K)
{
// Stores the absolute differences
// of adjacent elements
int[] new_Arr = new int[N - 1];
// Stores the answer
int ans = 0;
// Stores absolute differences of
// adjacent elements in new_Arr
for(int i = 0; i < N - 1; i++)
{
new_Arr[i] = Math.Abs(arr[i] -
arr[i + 1]);
// Stores the sum of all absolute
// differences of adjacent elements
ans += new_Arr[i];
}
// Sorting the new_Arr
// in decreasing order
Array.Sort(new_Arr);
reverse(new_Arr);
for(int i = 0; i < K - 1; i++)
{
// Removing K - 1 elements
// with maximum sum
ans -= new_Arr[i];
}
// Prints the answer
Console.WriteLine(ans);
}
// Driver Code
public static void Main ()
{
// Given array arr[]
int[] arr = { 1, 3, -2, 5, -1 };
// Given K
int K = 2;
// Size of array
int N = arr.Length;
// Function Call
absoluteDifference(arr, N, K);
}
}
// This code is contributed by susmitakundugoaldanga输出：
13时间复杂度： O(N * Log(N))
辅助空间： O(N)