Maximize difference between sum of even and odd-indexed elements of a subsequence

Given an array arr[] consisting of N positive integers, the task is to find the maximum possible difference between the sum of even and odd-indexed elements of a subsequence from the given array.

Examples:

Input: arr[] = { 3, 2, 1, 4, 5, 2, 1, 7, 8, 9 } 
Output: 15 
Explanation: 
Considering the subsequences { 3, 1, 5, 1, 9 } from the array 
Sum of even-indexed array elements = 3 + 5 + 9 = 17 
Sum of odd-indexed array elements = is 1 + 1 = 2 
Therefore, the difference between the sum of even and odd-indexed elements present in the subsequence = (17 - 2) = 15, which is the maximum possible.

Input: arr[] = {1, 2, 3, 4, 5, 6}
Output: 6

Naive Approach: The simplest approach to solve this problem is to generate all possible subsequences of the given array and for each subsequence, calculate the difference between the sum of even and odd indexed elements of the subsequence. Finally, print the maximum difference obtained. 

Time Complexity: O(2^N)
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, the idea is to store the local maxima at even indices of the subsequence and store the local minima at odd indices of the subsequence. Finally, print the difference between the sum of even and odd indices of the subsequence. 
Follow the steps below to solve the problem:

• Initialize a variable, say maxDiff, to store the maximum difference between the sum of even and odd-indexed elements of a subsequence.
• Traverse the array arr[] and check if arr[i] > arr[i + 1] and arr[i] < arr[i - 1] or not. If found to be true, then update maxDiff += arr[i].
• Otherwise, check if arr[i] > arr[i + 1] and arr[i] < arr[i - 1] or not. If found to be true, then update maxDiff -= arr[i].
• Finally, print the value of maxDiff.

Below is the implementation of the above approach:

// C++ program to implement
// the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to find the maximum possible difference
// between sum of even and odd indices
int maxPossibleDiff(vector<int>& arr, int N)
{

    // Convert arr[] into 1-based indexing
    arr.push_back(-1);

    // Reverse the array
    reverse(arr.begin(), arr.end());

    // Convert arr[] into 1 based index
    arr.push_back(-1);

    // Reverse the array
    reverse(arr.begin(), arr.end());

    // Stores maximum difference between
    // sum of even and odd indexed elements
    int maxDiff = 0;

    // Traverse the array
    for (int i = 1; i <= N; i++) {

        // If arr[i] is local maxima
        if (arr[i] > arr[i - 1]
            && arr[i] > arr[i + 1]) {

            // Update maxDiff
            maxDiff += arr[i];
        }

        // If arr[i] is local minima
        if (arr[i] < arr[i - 1]
            && arr[i] < arr[i + 1]) {

            // Update maxDiff
            maxDiff -= arr[i];
        }
    }
    cout << maxDiff;
}

// Driver Code
int main()
{

    vector<int> arr = { 3, 2, 1, 4, 5,
                        2, 1, 7, 8, 9 };

    // Size of array
    int N = arr.size();

    // Function Call
    maxPossibleDiff(arr, N);

    return 0;
}

// Java program to implement
// the above approach
import java.util.*;

class GFG{

// Function to find the maximum possible
// difference between sum of even and 
// odd indices
static void maxPossibleDiff(Vector<Integer> arr, int N)
{
    
    // Convert arr[] into 1-based indexing
    arr.add(-1);

    // Reverse the array
    Collections.reverse(arr);

    // Convert arr[] into 1 based index
    arr.add(-1);

    // Reverse the array
    Collections.reverse(arr);

    // Stores maximum difference between
    // sum of even and odd indexed elements
    int maxDiff = 0;

    // Traverse the array
    for(int i = 1; i <= N; i++) 
    {
        
        // If arr.get(i) is local maxima
        if (arr.get(i) > arr.get(i - 1) && 
            arr.get(i) > arr.get(i + 1)) 
        {
            
            // Update maxDiff
            maxDiff += arr.get(i);
        }

        // If arr.get(i) is local minima
        if (arr.get(i) < arr.get(i - 1) && 
            arr.get(i) < arr.get(i + 1))
        {
            
            // Update maxDiff
            maxDiff -= arr.get(i);
        }
    }
    System.out.print(maxDiff);
}

// Driver Code
public static void main(String[] args)
{
    int[] array = { 3, 2, 1, 4, 5,
                    2, 1, 7, 8, 9 };
    Vector<Integer> v = new Vector<>();
    for(int i :array) 
    {
        v.add(i);
    }
    
    // Size of array
    int N = v.size();

    // Function Call
    maxPossibleDiff(v, N);
}
}

// This code is contributed by shikhasingrajput

#Python3 program to implement
#the above approach


#Function to find the maximum possible difference
#between sum of even and odd indices
def maxPossibleDiff(arr,  N):

    #Convert arr[] o 1-based indexing
    arr.append(-1)

    #Reverse the array
    arr = arr[::-1]

    #Convert arr[] o 1 based index
    arr.append(-1)

    #Reverse the array
    arr = arr[::-1]

    #Stores maximum difference between
    #sum of even and odd indexed elements
    maxDiff = 0

    #Traverse the array
    for i in range(1,N+1):

        #If arr[i] is local maxima
        if (arr[i] > arr[i - 1] and arr[i] > arr[i + 1]):

            #Update maxDiff
            maxDiff += arr[i]

        #If arr[i] is local minima
        if (arr[i] < arr[i - 1] and arr[i] < arr[i + 1]):

            #Update maxDiff
            maxDiff -= arr[i]
    print (maxDiff)

#Driver Code
if __name__ == '__main__':

    arr = [3, 2, 1, 4, 5, 2, 1, 7, 8, 9]

    #Size of array
    N = len(arr)

    #Function Call
    maxPossibleDiff(arr, N)

// C# program to implement
// the above approach
using System;
using System.Collections.Generic;

public class GFG{

// Function to find the maximum possible
// difference between sum of even and 
// odd indices
static void maxPossibleDiff(List<int> arr, int N)
{
    
    // Convert []arr into 1-based indexing
    arr.Add(-1);

    // Reverse the array
    arr.Reverse();

    // Convert []arr into 1 based index
    arr.Add(-1);

    // Reverse the array
    arr.Reverse();

    // Stores maximum difference between
    // sum of even and odd indexed elements
    int maxDiff = 0;

    // Traverse the array
    for(int i = 1; i <= N; i++) 
    {
        
        // If arr[i] is local maxima
        if (arr[i] > arr[i - 1] && 
            arr[i] > arr[i + 1]) 
        {
            
            // Update maxDiff
            maxDiff += arr[i];
        }

        // If arr[i] is local minima
        if (arr[i] < arr[i - 1] && 
            arr[i] < arr[i + 1])
        {
            
            // Update maxDiff
            maxDiff -= arr[i];
        }
    }
    Console.Write(maxDiff);
}

// Driver Code
public static void Main(String[] args)
{
    int[] array = { 3, 2, 1, 4, 5,
                    2, 1, 7, 8, 9 };
    List<int> v = new List<int>();
    foreach(int i in array) 
    {
        v.Add(i);
    }
    
    // Size of array
    int N = v.Count;

    // Function Call
    maxPossibleDiff(v, N);
}
}

// This code is contributed by 29AjayKumar 

<script>
 
// JavaScript program to implement
// the above approach

// Function to find the maximum possible difference
// between sum of even and odd indices
function maxPossibleDiff(arr, N)
{

    // Convert arr[] into 1-based indexing
    arr.push(-1);

    // Reverse the array
    arr.reverse();

    // Convert arr[] into 1 based index
    arr.push(-1);

    // Reverse the array
    arr.reverse();

    // Stores maximum difference between
    // sum of even and odd indexed elements
    var maxDiff = 0;

    // Traverse the array
    for (var i = 1; i <= N; i++) {

        // If arr[i] is local maxima
        if (arr[i] > arr[i - 1]
            && arr[i] > arr[i + 1]) {

            // Update maxDiff
            maxDiff += arr[i];
        }

        // If arr[i] is local minima
        if (arr[i] < arr[i - 1]
            && arr[i] < arr[i + 1]) {

            // Update maxDiff
            maxDiff -= arr[i];
        }
    }
    document.write( maxDiff);
}

// Driver Code

var arr = [3, 2, 1, 4, 5,
                    2, 1, 7, 8, 9];
                    
// Size of array
var N = arr.length;

// Function Call
maxPossibleDiff(arr, N);

</script>    
    

15

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