Smallest subsequence with sum of absolute difference of consecutive elements maximized

Given an array arr[] of length N, containing values in the range [1, N], the task is to find a subsequence s₁, s₂, …, s_k such that 

is maximised. In case of multiple subsequences having maximum possible sum, print the smallest one.

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. 

Naive Approach:
Generate all subsequences of length >= 2 and calculate their respective sums. Keep track of the maximum sum obtained for any subsequence. In the case of multiple subsequences having the maximum sum, keep updating the minimum length, and maintain that subsequence. Finally print the subsequence. 
Time Complexity: O(2^N)
Efficient Approach:
There are a few key observations that will help us proceed: 

1. The maximum sum will be obtained when all the elements in the permutation are considered. 
   For example: 
   

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. 

1. 
   Now that we know the maximum possible sum, the objective is to minimize the subsequence length without affecting this maximum sum.
2. For a monotonically increasing or decreasing subsequence, maximum sum can be achieved by only considering the First and Last element, for example: 
   

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 

In this way, the length of the subsequence can be reduced without affecting the Max sum. 
Hence, from the given array, keep extracting the end-points of monotonically increasing or decreasing subsequences, and add them to the subsequence in the following way: 

• The first and the last element of the permutation are default endpoints

• An element P[ i ] is a monotonically increasing endpoint if P[ i – 1 ] < P[ i ] > P[ i + 1 ]

• An element P[ i ] is monotonically decreasing endpoint if P[ i – 1 ] > P[ i ] < P[ i + 1 ]

The subsequence thus obtained will have maximum sum and minimum length.
Below is the implementation of the above approach:

C++

// C++ program to find
// smallest subsequence
// with sum of absolute
// difference of consecutive
// elements maximized
#include <bits/stdc++.h>
using namespace std;

// Function to print the smallest
// subsequence and its sum
void getSubsequence(vector<int>& arr,
                    int n)
{
    // Final subsequence
    vector<int> 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<int> 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<Integer> req = new Vector<Integer>();

    // 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<int> req = new List<int>();

    // 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

Javascript

<script>
        // JavaScript code for the above approach

        // Function to print the smallest
        // subsequence and its sum
        function getSubsequence(arr, n)
        {

            // Final subsequence
            let req = [];

            // First element is
            // a default endpoint
            req.push(arr[0]);

            // Iterating through the array
            for (let i = 1; i < n - 1; i++) {

                // Check for monotonically
                // increasing endpoint
                if (arr[i] > arr[i + 1]
                    && arr[i] > arr[i - 1])
                    req.push(arr[i]);

                // Check for monotonically
                // decreasing endpoint
                else if (arr[i] < arr[i + 1]
                    && arr[i] < arr[i - 1])
                    req.push(arr[i]);
            }

            // Last element is
            // a default endpoint
            req.push(arr[n - 1]);

            // Length of final subsequence
            document.write(req.length + '<br>');

            // Print the subsequence
            for (let x of req)
                document.write(x + " ");
        }

        // Driver Program
        let arr = [1, 2, 5, 3,
            6, 7, 4];
        let n = arr.length;
        getSubsequence(arr, n);

  // This code is contributed by Potta Lokesh
    </script>

Output:

5
1 5 3 7 4

Time complexity: O(N)