最小子序列，连续元素的绝对差之和最大

给定长度为N的数组arr [] ，其中包含范围为[1，N]的值，任务是找到子序列s ₁ ，s ₂ ，…，s _k ，使得
$\\ Sum = \mid s1 - s2 \mid + \mid s2 - s3 \mid + ... + \mid s_{k-1} - s_{k} \mid$
被最大化。如果多个子序列的总和最大，则打印最小的子序列。

Input: N = 3, P = {3, 2, 1}
Output: K = 2, S = {3, 1}
Explanation:
Maximum sum possible = 2
Subsequences {3, 2, 1} and {3, 1} achieves that sum.
Hence, the subsequence {3, 1} is considered being of smaller length.

Input: N = 4, P = {1, 3, 4, 2}
Output: K = 3, S = {1, 4, 2}
Explanation:
Maximum sum possible = 5
Subsequences {1, 3, 4, 2} and {1, 4, 2} achieves that sum.
Hence, the subsequence {1, 4, 2} is considered being of smaller length.

天真的方法：
生成所有长度> = 2的子序列，并计算它们各自的总和。跟踪获得的任何子序列的最大和。在多个子序列的总和最大的情况下，请继续更新最小长度，并保持该子序列的大小。最后打印子序列。
时间复杂度： O(2 ^N )

高效方法：
有一些关键的观察将有助于我们继续前进：

1. 当考虑置换中的所有元素时，将获得最大和。
   例如：

   N = 4, P = [1, 3, 4, 2]
   Max sum = | 1 – 3 | + | 3 – 4 | + | 4 – 2 | = 2 + 1 + 2 = 5
   Here, the length is N and the required subsequence is permutation P itself.

   既然我们知道了最大可能的和，我们的目标是在不影响该最大和的情况下最小化子序列的长度。

2. 对于单调增加或减少的子序列，仅考虑First和Last元素即可达到最大总和，例如：

   S = [1, 3, 5, 7]
   Max sum = | 1 – 3 | + | 3 – 5 | + | 5 – 7 | = 2 + 2 + 2 = 6, K = 4
   Considering only first and last element,
   S = [1, 7]
   Max sum = | 1 – 7 | = 6, K = 2

这样，可以减小子序列的长度而不会影响Max sum 。

因此，从给定数组中，继续提取单调递增或递减子序列的端点，并通过以下方式将其添加到子序列中：

排列的第一个和最后一个元素是默认端点

如果P [i – 1]

P [i + 1]，则元素P [i]是单调递增的端点。

如果P [i – 1]> P [i]

元素P [i]是单调递减的端点

这样获得的子序列将具有最大和和最小长度。

下面是上述方法的实现：

C++

// C++ program to find
// smallest subsequence
// with sum of absolute
// difference of consecutive
// elements maximized
#include 
using namespace std;
  
// Function to print the smallest
// subsequence and its sum
void getSubsequence(vector& arr,
                    int n)
{
    // Final subsequence
    vector req;
  
    // First element is
    // a default endpoint
    req.push_back(arr[0]);
  
    // Iterating through the array
    for (int i = 1; i < n - 1; i++) {
  
        // Check for monotonically
        // increasing endpoint
        if (arr[i] > arr[i + 1]
            && arr[i] > arr[i - 1])
            req.push_back(arr[i]);
  
        // Check for monotonically
        // decreasing endpoint
        else if (arr[i] < arr[i + 1]
                 && arr[i] < arr[i - 1])
            req.push_back(arr[i]);
    }
  
    // Last element is
    // a default endpoint
    req.push_back(arr[n - 1]);
  
    // Length of final subsequence
    cout << req.size() << endl;
  
    // Print the subsequence
    for (auto x : req)
        cout << x << " ";
}
  
// Driver Program
int main()
{
    vector arr = { 1, 2, 5, 3,
                        6, 7, 4 };
    int n = arr.size();
    getSubsequence(arr, n);
  
    return 0;
}

Java

// Java program to find smallest
// subsequence with sum of absolute
// difference of consecutive
// elements maximized
import java.util.*;
  
class GFG{
  
// Function to print the smallest
// subsequence and its sum
static void getSubsequence(int []arr, int n)
{
      
    // Final subsequence
    Vector req = new Vector();
  
    // First element is
    // a default endpoint
    req.add(arr[0]);
  
    // Iterating through the array
    for(int i = 1; i < n - 1; i++)
    {
          
       // Check for monotonically
       // increasing endpoint
       if (arr[i] > arr[i + 1] && 
           arr[i] > arr[i - 1])
           req.add(arr[i]);
             
       // Check for monotonically
       // decreasing endpoint
       else if (arr[i] < arr[i + 1] && 
                arr[i] < arr[i - 1])
           req.add(arr[i]);
    }
  
    // Last element is
    // a default endpoint
    req.add(arr[n - 1]);
  
    // Length of final subsequence
    System.out.print(req.size() + "\n");
  
    // Print the subsequence
    for(int x : req)
       System.out.print(x + " ");
}
  
// Driver code
public static void main(String[] args)
{
    int []arr = { 1, 2, 5, 3,
                  6, 7, 4 };
    int n = arr.length;
      
    getSubsequence(arr, n);
}
}
  
// This code is contributed by Amit Katiyar

Python3

# Python3 program to find smallest
# subsequence with sum of absolute
# difference of consecutive
# elements maximized
  
# Function to print the smallest
# subsequence and its sum
def getSubsequence(arr, n):
      
    # Final subsequence
    req = []
      
    # First element is
    # a default endpoint
    req.append(arr[0])
      
    # Iterating through the array
    for i in range(1, n - 1):
          
        # Check for monotonically
        # increasing endpoint
        if (arr[i] > arr[i + 1] and
            arr[i] > arr[i - 1]):
            req.append(arr[i])
         
        # Check for monotonically
        # decreasing endpoint
        elif (arr[i] < arr[i + 1] and 
              arr[i] < arr[i - 1]):
            req.append(arr[i]);
              
    # Last element is
    # a default endpoint
    req.append(arr[n - 1]);
  
    # Length of final subsequence
    print(len(req))
  
    # Print the subsequence
    for x in req:
        print(x, end = ' ')
  
# Driver code
if __name__=='__main__':
      
    arr = [ 1, 2, 5, 3, 6, 7, 4 ]
    n = len(arr)
      
    getSubsequence(arr, n)
  
# This code is contributed by rutvik_56

C#

// C# program to find smallest
// subsequence with sum of absolute
// difference of consecutive
// elements maximized
using System;
using System.Collections.Generic;
  
class GFG{
   
// Function to print the smallest
// subsequence and its sum
static void getSubsequence(int []arr, int n)
{
       
    // Final subsequence
    List req = new List();
   
    // First element is
    // a default endpoint
    req.Add(arr[0]);
   
    // Iterating through the array
    for(int i = 1; i < n - 1; i++)
    {
           
       // Check for monotonically
       // increasing endpoint
       if (arr[i] > arr[i + 1] && 
           arr[i] > arr[i - 1])
           req.Add(arr[i]);
              
       // Check for monotonically
       // decreasing endpoint
       else if (arr[i] < arr[i + 1] && 
                arr[i] < arr[i - 1])
           req.Add(arr[i]);
    }
   
    // Last element is
    // a default endpoint
    req.Add(arr[n - 1]);
   
    // Length of readonly subsequence
    Console.Write(req.Count + "\n");
   
    // Print the subsequence
    foreach(int x in req)
       Console.Write(x + " ");
}
   
// Driver code
public static void Main(String[] args)
{
    int []arr = { 1, 2, 5, 3,
                  6, 7, 4 };
    int n = arr.Length;
       
    getSubsequence(arr, n);
}
}
  
// This code is contributed by Amit Katiyar

输出：

5
1 5 3 7 4

时间复杂度： O(N)

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