Given an array arr[] of length N, containing values in the range [1, N], the task is to find a subsequence s1, s2, …, sk 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(2N)
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 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 sumvoid 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 Programint 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 maximizedimport java.util.*;class GFG{// Function to print the smallest// subsequence and its sumstatic 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 codepublic 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 sumdef 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 codeif __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 maximizedusing System;using System.Collections.Generic;class GFG{ // Function to print the smallest// subsequence and its sumstatic 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 codepublic 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> |
5 1 5 3 7 4
Time complexity: O(N)
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