Subsequence with maximum pairwise absolute difference and minimum size

Given an array arr[] of N integers, the task is to print the subsequence from the given array with the maximum pairwise absolute difference of adjacent elements. If multiple such subsequences exist, then print the subsequence with the smallest length.
Examples: 
 

Input: arr[] = {1, 2, 4, 3, 5} 
Output: 1 4 3 5 
Explanation: 
For the subsequence {1, 4, 3, 5}, 
The absolute difference between adjacent pair is |1 – 4| + |4 – 3| + |3 – 5| = 6, which is the maximum possible absolute difference.
Input: arr[] = {1, 2, 5, 6, 3, 4} 
Output: {1, 6, 4} 
Explanation: 
For the subsequence {1, 6, 4}, 
The absolute difference between adjacent pair is |1 – 6| + |6 – 4| = 7, which is the maximum possible absolute difference. 
 

 

Approach: For the subsequence with the maximum absolute difference, below are the steps: 
 

  1. Create the new array(say ans[]) to store all the element of the required subsequence.
  2. Insert the first element of the array arr[].
  3. Find all the local minima and maxima of the array excluding the first and last element of the array and insert in the array ans[].
  4. Insert the last element of the array arr[].

Below is the implementation of the above approach:
 



C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the subsequence
// with maximum absolute difference
void getSubsequence(vector<int> ar)
{
    int N = ar.size();
  
    // To store the resultant subsequence
    vector<int> ans;
  
    // First element should be included
    // in the subsequence
    ans.push_back(ar[0]);
  
    // Traverse the given array arr[]
    for (int i = 1; i < N - 1; i++) {
  
        // If current element is greater
        // than the previous element
        if (ar[i] > ar[i - 1]) {
  
            // If the current element is
            // not the local maxima
            // then continue
            if (i < N - 1
                && ar[i] <= ar[i + 1]) {
                continue;
            }
  
            // Else push it in subsequence
            else {
                ans.push_back(ar[i]);
            }
        }
  
        // If the current element is less
        // then the previous element
        else {
  
            // If the current element is
            // not the local minima
            // then continue
            if (i < N - 1
                && ar[i + 1] < ar[i]) {
                continue;
            }
  
            // Else push it in subsequence
            else {
                ans.push_back(ar[i]);
            }
        }
    }
  
    // Last element should also be
    // included in subsequence
    ans.push_back(ar[N - 1]);
  
    // Print the element
    for (auto& it : ans)
        cout << it << ' ';
}
  
// Driver Code
int main()
{
    // Given array
    vector<int> arr = { 1, 2, 4, 3, 5 };
  
    // Function Call
    getSubsequence(arr);
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program for the above approach
import java.util.*;
  
class GFG{
  
// Function to find the subsequence
// with maximum absolute difference
static void getSubsequence(int []ar)
{
    int N = ar.length;
  
    // To store the resultant subsequence
    Vector<Integer> ans = new Vector<Integer>();
  
    // First element should be included
    // in the subsequence
    ans.add(ar[0]);
  
    // Traverse the given array arr[]
    for(int i = 1; i < N - 1; i++)
    {
         
       // If current element is greater
       // than the previous element
       if (ar[i] > ar[i - 1])
       {
             
           // If the current element is
           // not the local maxima
           // then continue
           if (i < N - 1 && ar[i] <= ar[i + 1])
           {
               continue;
           }
             
           // Else push it in subsequence
           else
           {
               ans.add(ar[i]);
           }
       }
         
       // If the current element is less
       // then the previous element
       else
       {
             
           // If the current element is
           // not the local minima
           // then continue
           if (i < N - 1 && ar[i + 1] < ar[i])
           {
               continue;
           }
             
           // Else push it in subsequence
           else
           {
               ans.add(ar[i]);
           }
       }
    }
  
    // Last element should also be
    // included in subsequence
    ans.add(ar[N - 1]);
  
    // Print the element
    for(int it : ans)
       System.out.print(it + " ");
}
  
// Driver Code
public static void main(String[] args)
{
      
    // Given array
    int []arr = { 1, 2, 4, 3, 5 };
  
    // Function Call
    getSubsequence(arr);
}
}
  
// This code is contributed by Princi Singh

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 program for the above approach
  
# Function to find the subsequence
# with maximum absolute difference
def getSubsequence(ar):
  
    N = len(ar)
  
    # To store the resultant subsequence
    ans = []
  
    # First element should be included
    # in the subsequence
    ans.append(ar[0])
  
    # Traverse the given array arr[]
    for i in range(1, N - 1):
  
        # If current element is greater
        # than the previous element
        if (ar[i] > ar[i - 1]):
  
            # If the current element is
            # not the local maxima
            # then continue
            if(i < N - 1 and 
               ar[i] <= ar[i + 1]):
                continue
  
            # Else push it in subsequence
            else:
                ans.append(ar[i])
  
        # If the current element is less
        # then the previous element
        else:
  
            # If the current element is
            # not the local minima
            # then continue
            if (i < N - 1 and 
                 ar[i + 1] < ar[i]):
                continue
  
                # Else push it in subsequence
            else:
                ans.append(ar[i])
  
    # Last element should also be
    # included in subsequence
    ans.append(ar[N - 1])
  
    # Print the element
    for it in ans:
        print(it, end = " ")
  
# Driver Code
if __name__ == '__main__':
  
    # Given array
    arr = [ 1, 2, 4, 3, 5 ]
  
    # Function Call
    getSubsequence(arr)
  
# This code is contributed by Shivam Singh

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program for the above approach 
using System;
using System.Collections.Generic;
  
class GFG{ 
  
// Function to find the subsequence 
// with maximum absolute difference 
static void getSubsequence(int []ar) 
    int N = ar.Length; 
  
    // To store the resultant subsequence 
    List<int> ans = new List<int>(); 
  
    // First element should be included 
    // in the subsequence 
    ans.Add(ar[0]); 
  
    // Traverse the given array []arr 
    for(int i = 1; i < N - 1; i++) 
    
          
        // If current element is greater 
        // than the previous element 
        if (ar[i] > ar[i - 1]) 
        
              
            // If the current element is 
            // not the local maxima 
            // then continue 
            if (i < N - 1 && ar[i] <= ar[i + 1]) 
            
                continue
            
                  
            // Else push it in subsequence 
            else
            
                ans.Add(ar[i]); 
            
        
              
        // If the current element is less 
        // then the previous element 
        else
        
              
            // If the current element is 
            // not the local minima 
            // then continue 
            if (i < N - 1 && ar[i + 1] < ar[i]) 
            
                continue
            
                  
            // Else push it in subsequence 
            else
            
                ans.Add(ar[i]); 
            
        
    
      
    // Last element should also be 
    // included in subsequence 
    ans.Add(ar[N - 1]); 
  
    // Print the element 
    foreach(int it in ans) 
        Console.Write(it + " "); 
  
// Driver Code 
public static void Main(String[] args) 
      
    // Given array 
    int []arr = { 1, 2, 4, 3, 5 }; 
  
    // Function Call 
    getSubsequence(arr); 
  
// This code is contributed by Princi Singh

chevron_right


Output: 

1 4 3 5

 

Time Complexity: O(N) 
Auxiliary Space: O(1)
 

competitive-programming-img




My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.