Given an array **arr[]** of length **N**, containing values in the range **[1, N]**, the task is to find a subsequence **s _{1}, s_{2}, …, 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:

- 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.Now that we know the maximum possible sum, the objective is to minimize the subsequence length without affecting this maximum sum.

- For a monotonically increasing or decreasing subsequence, maximum sum can be acheived 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:

**first**and the

**last**element of the permutation are default endpoints

**P[ i ]**is a

**monotonically increasing endpoint**if

**P[ i – 1 ] < P[ i ] > P[ i + 1 ]**

**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` |

**Output:**

5 1 5 3 7 4

**Time complexity:** O(N)