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Maximize difference between sum of even and odd-indexed elements of a subsequence | Set 2

  • Last Updated : 17 Jan, 2022

Given an array arr[] consisting of N positive integers, the task is to find the maximum value of the difference between the sum of elements at even and odd indices for any subsequence of the array.

Note: The value of N is always greater than 1.

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 simple approach to solve the given problem is to generate all possible subsequences of the given array and for each subsequence maximize the difference between the sum of elements at even and odd indices. After checking for all subsequences, print the maximum value obtained.

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

Local Maxima Approach: The approach to solve this problem using Local Maxima and Local Minima is discussed in this post. In this article, we have discussed the dynamic programming approach.

Efficient Approach: The above approach can also be optimized by using Dynamic Programming as the above approach has Optimal Substructure and Overlapping Subproblems. Follow the steps below to solve the given problem:

  • Initialize two arrays, dp1[] and dp2[] of size N, and initialize with values -1 such that dp1[i] stores the maximum sum from a subsequence of the odd length till the ith index and dp2[i] stores the maximum sum from a subsequence of an even length till the ith index.
  • Update the value of dp1[0] as arr[0] and dp2[0] as 0.
  • Iterate over the range [1, N] using the variable i and perform the following steps:
    • Update the value of dp1[i] as dp1[i] = max(dp1[i – 1], dp2[i – 1] + arr[i]), as the ith element will be appended at an even index in the subsequence. Hence, the array element arr[i] is added.
    • Update the value of dp2[i] as dp2[i] = max(dp2[i – 1], dp1[i – 1] – arr[i]) as the ith element will be appended at an odd index in the subsequence. Hence, the array element arr[i] is subtracted.
  • After performing the above steps, print the maximum of (dp1[N – 1], dp2[N – 1]) as the result.

Below is the implementation of the above approach:

C++14




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the maximum value
// of difference between the sum of
// elements at even and odd indices of
// any subsequence of the array
void findMaximumPeakSum(int arr[], int n)
{
    // Initialize the two arrays
    int dp1[n], dp2[n];
    for (int i = 0; i < n; i++) {
        dp1[i] = -1;
        dp2[i] = -1;
    }
    dp2[0] = 0;
    dp1[0] = arr[0];
 
    // Iterate over the range
    for (int i = 1; i < n; i++) {
 
        // Find the maximum sum upto the
        // i-th odd and even subsequence
        dp1[i] = max(dp1[i - 1],
                     dp2[i - 1] + arr[i]);
        dp2[i] = max(dp2[i - 1],
                     dp1[i - 1] - arr[i]);
    }
 
    // Find the maximum value
    int ans = max(dp1[n - 1], dp2[n - 1]);
 
    cout << ans;
}
 
// Driver Code
int main()
{
    int arr[] = { 3, 2, 1, 4, 5, 2, 1, 7, 8, 9 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    findMaximumPeakSum(arr, N);
 
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to find the maximum value
// of difference between the sum of
// elements at even and odd indices of
// any subsequence of the array
static void findMaximumPeakSum(int arr[], int n)
{
   
    // Initialize the two arrays
    int []dp1 = new int[n];
    int []dp2 = new int[n];
    for (int i = 0; i < n; i++) {
        dp1[i] = -1;
        dp2[i] = -1;
    }
    dp2[0] = 0;
    dp1[0] = arr[0];
 
    // Iterate over the range
    for (int i = 1; i < n; i++) {
 
        // Find the maximum sum upto the
        // i-th odd and even subsequence
        dp1[i] = Math.max(dp1[i - 1],
                     dp2[i - 1] + arr[i]);
        dp2[i] = Math.max(dp2[i - 1],
                     dp1[i - 1] - arr[i]);
    }
 
    // Find the maximum value
    int ans = Math.max(dp1[n - 1], dp2[n - 1]);
 
    System.out.print(ans);
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 3, 2, 1, 4, 5, 2, 1, 7, 8, 9 };
    int N = arr.length;
 
    findMaximumPeakSum(arr, N);
}
}
 
// This code is contributed by shikhasingrajput

Python3




# Python Program to implement
# the above approach
 
# Function to find the maximum value
# of difference between the sum of
# elements at even and odd indices of
# any subsequence of the array
def findMaximumPeakSum(arr, n):
   
    # Initialize the two arrays
    dp1 = [0] * n
    dp2 = [0] * n
    for i in range(n):
        dp1[i] = -1
        dp2[i] = -1
 
    dp2[0] = 0
    dp1[0] = arr[0]
 
    # Iterate over the range
    for i in range(1, n):
 
        # Find the maximum sum upto the
        # i-th odd and even subsequence
        dp1[i] = max(dp1[i - 1],
                     dp2[i - 1] + arr[i])
        dp2[i] = max(dp2[i - 1],
                     dp1[i - 1] - arr[i])
 
    # Find the maximum value
    ans = max(dp1[n - 1], dp2[n - 1])
 
    print(ans)
 
# Driver Code
arr = [3, 2, 1, 4, 5, 2, 1, 7, 8, 9]
N = len(arr)
 
findMaximumPeakSum(arr, N)
 
# This code is contributed by Saurabh Jaiswal

C#




// C# program for the above approach
using System;
 
public class GFG{
 
// Function to find the maximum value
// of difference between the sum of
// elements at even and odd indices of
// any subsequence of the array
static void findMaximumPeakSum(int []arr, int n)
{
   
    // Initialize the two arrays
    int []dp1 = new int[n];
    int []dp2 = new int[n];
    for (int i = 0; i < n; i++) {
        dp1[i] = -1;
        dp2[i] = -1;
    }
    dp2[0] = 0;
    dp1[0] = arr[0];
 
    // Iterate over the range
    for (int i = 1; i < n; i++) {
 
        // Find the maximum sum upto the
        // i-th odd and even subsequence
        dp1[i] = Math.Max(dp1[i - 1],
                     dp2[i - 1] + arr[i]);
        dp2[i] = Math.Max(dp2[i - 1],
                     dp1[i - 1] - arr[i]);
    }
 
    // Find the maximum value
    int ans = Math.Max(dp1[n - 1], dp2[n - 1]);
 
    Console.Write(ans);
}
 
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 3, 2, 1, 4, 5, 2, 1, 7, 8, 9 };
    int N = arr.Length;
 
    findMaximumPeakSum(arr, N);
}
}
 
// This code is contributed by shikhasingrajput

Javascript




<script>
       // JavaScript Program to implement
       // the above approach
 
       // Function to find the maximum value
       // of difference between the sum of
       // elements at even and odd indices of
       // any subsequence of the array
       function findMaximumPeakSum(arr, n) {
           // Initialize the two arrays
           let dp1 = new Array(n);
           let dp2 = new Array(n);
           for (let i = 0; i < n; i++) {
               dp1[i] = -1;
               dp2[i] = -1;
           }
           dp2[0] = 0;
           dp1[0] = arr[0];
 
           // Iterate over the range
           for (let i = 1; i < n; i++) {
 
               // Find the maximum sum upto the
               // i-th odd and even subsequence
               dp1[i] = Math.max(dp1[i - 1],
                   dp2[i - 1] + arr[i]);
               dp2[i] = Math.max(dp2[i - 1],
                   dp1[i - 1] - arr[i]);
           }
 
           // Find the maximum value
           let ans = Math.max(dp1[n - 1], dp2[n - 1]);
 
           document.write(ans);
       }
 
       // Driver Code
       let arr = [3, 2, 1, 4, 5, 2, 1, 7, 8, 9];
       let N = arr.length;
 
       findMaximumPeakSum(arr, N);
 
   // This code is contributed by Potta Lokesh
   </script>
Output
15

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

Optimized Approach: The above approach can be further be optimized by using two variables, odd and even, instead of two arrays, dp1[] and dp2[] to maintain the maximum difference between the sum of elements at even and odd indices. For every index i, only the maximum sums of even and odd length subsequences of the previous index are required to calculate the current states are needed.

Below is the implementation of the above approach: 

C++14




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the maximum value
// of difference between the sum of
// elements at even and odd indices of
// any subsequence of the array
void findMaximumPeakSum(int arr[], int n)
{
    // Initialize the variables
    int even = 0;
    int odd = arr[0];
 
    // Iterat over the range
    for (int i = 1; i < n; i++) {
 
        // Find the maximum sum upto the
        // i-th odd and even subsequence
        int temp = odd;
        odd = max(odd, even + arr[i]);
        even = max(even, temp - arr[i]);
    }
 
    // Find the resultant maximum value
    int ans = max(odd, even);
 
    cout << ans;
}
 
// Driver Code
int main()
{
    int arr[] = { 3, 2, 1, 4, 5, 2, 1, 7, 8, 9 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    findMaximumPeakSum(arr, N);
 
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
class GFG {
 
// Function to find the maximum value
// of difference between the sum of
// elements at even and odd indices of
// any subsequence of the array
static void findMaximumPeakSum(int arr[], int n)
{
   
    // Initialize the variables
    int even = 0;
    int odd = arr[0];
  
    // Iterat over the range
    for (int i = 1; i < n; i++) {
  
        // Find the maximum sum upto the
        // i-th odd and even subsequence
        int temp = odd;
        odd = Math.max(odd, even + arr[i]);
        even = Math.max(even, temp - arr[i]);
    }
  
    // Find the resultant maximum value
    int ans = Math.max(odd, even);
  
    System.out.print(ans);
}
  
    // Driver Code
    public static void main(String[] args) {
        int arr[] = { 3, 2, 1, 4, 5, 2, 1, 7, 8, 9 };
    int N = arr.length;
  
    findMaximumPeakSum(arr, N);
    }
}
 
// This code is contributed by code_hunt.

Python3




# Python 3 program for the above approach
 
# Function to find the maximum value
# of difference between the sum of
# elements at even and odd indices of
# any subsequence of the array
def findMaximumPeakSum(arr, n):
   
    # Initialize the variables
    even = 0
    odd = arr[0]
 
    # Iterat over the range
    for i in range(1,  n):
 
        # Find the maximum sum upto the
        # i-th odd and even subsequence
        temp = odd
        odd = max(odd, even + arr[i])
        even = max(even, temp - arr[i])
 
    # Find the resultant maximum value
    ans = max(odd, even)
 
    print(ans)
 
# Driver Code
if __name__ == "__main__":
 
    arr = [3, 2, 1, 4, 5, 2, 1, 7, 8, 9]
    N = len(arr)
 
    findMaximumPeakSum(arr, N)
 
    # This code is contributed by ukasp.

C#




// C# program for the above approach
using System;
class GFG {
 
// Function to find the maximum value
// of difference between the sum of
// elements at even and odd indices of
// any subsequence of the array
static void findMaximumPeakSum(int[] arr, int n)
{
   
    // Initialize the variables
    int even = 0;
    int odd = arr[0];
  
    // Iterat over the range
    for (int i = 1; i < n; i++) {
  
        // Find the maximum sum upto the
        // i-th odd and even subsequence
        int temp = odd;
        odd = Math.Max(odd, even + arr[i]);
        even = Math.Max(even, temp - arr[i]);
    }
  
    // Find the resultant maximum value
    int ans = Math.Max(odd, even);
  
    Console.Write(ans);
}
 
// Driver Code
public static void Main () {
         
    int[] arr = { 3, 2, 1, 4, 5, 2, 1, 7, 8, 9 };
    int N = arr.Length;
  
    findMaximumPeakSum(arr, N);
}
}
 
// This code is contributed by sanjoy_62.
Output
15

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

 


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