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Maximize subsequence sum after putting plus minus sign alternatively on elements

  • Last Updated : 07 Dec, 2021

Given an array arr[] of size N. The task is to find the maximum score that can be achieved by alternative minus – plus of elements of any subsequence in the array.

Example:

Input: arr[] = {2, 3, 6, 5, 4, 7, 8}, N = 7
Output: 10
Explanation: Pick the sequence as {2, 3, 6, 5, 4, 7, 8} = {6, 4, 8}. So, alternative minus – plus = (6 4 + 8) = 10.

Input: {9, 2, 4, 5, 3}, N =5
Output: 12

 

Approach: We can use Dynamic Programming method to solve the problem.

  • Let dp1i ⇢ be the maximum possible sum of a subsequence on a prefix from the first i elements, provided that the length of the subsequence is odd. Similarly enter dp2i ⇢ only for subsequences of even length.
  • Then dp1i and dp2i are easy to recalculate as :
    • dp1i+1 = max(dp1i, dp2i+arri)
    • dp2i+1 = max(dp2i, dp1i−arri)
  • The initial values are dp10 = −∞, dp20 = 0 and the answer will be stored in max(dp1n, dp2n).

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
#define INF 1e9
 
// Function to find the maximum score
// achieved by alternative minus-plus
// of elements of a subsequence in
// the given array
int maxScore(int arr[], int N)
{
    vector<int> dp1(N + 1);
    vector<int> dp2(N + 1);
 
    dp1[0] = -INF;
    dp2[0] = 0;
 
    for (int i = 0; i < N; ++i) {
        dp1[i + 1] = max(dp1[i], dp2[i] + arr[i]);
        dp2[i + 1] = max(dp2[i], dp1[i] - arr[i]);
    }
    // Return the maximum
    return max(dp1.back(), dp2.back());
}
 
// Driver Code
int main()
{
    int arr[] = { 2, 3, 6, 5, 4, 7, 8 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    cout << maxScore(arr, N);
    return 0;
}

Java




// Java program for the above approach
class GFG {
 
    static double INF = 1E9;
 
    // Function to find the maximum score
    // achieved by alternative minus-plus
    // of elements of a subsequence in
    // the given array
    public static int maxScore(int arr[], int N) {
        int[] dp1 = new int[N + 1];
        int[] dp2 = new int[N + 1];
 
        dp1[0] = (int) -INF;
        dp2[0] = 0;
 
        for (int i = 0; i < N; ++i) {
            dp1[i + 1] = Math.max(dp1[i], dp2[i] + arr[i]);
            dp2[i + 1] = Math.max(dp2[i], dp1[i] - arr[i]);
        }
       
        // Return the maximum
        return Math.max(dp1[dp1.length - 1], dp2[dp2.length - 1]);
    }
 
    // Driver Code
    public static void main(String args[]) {
        int arr[] = { 2, 3, 6, 5, 4, 7, 8 };
        int N = arr.length;
 
        System.out.println(maxScore(arr, N));
    }
}
 
// This code is contributed by saurabh_jaiswal.

Python3




# Python Program to implement
# the above approach
INF = 1e9
 
# Function to find the maximum score
# achieved by alternative minus-plus
# of elements of a subsequence in
# the given array
def maxScore(arr, N):
    dp1 = [0] * (N + 1)
    dp2 = [0] * (N + 1)
 
    dp1[0] = -INF
    dp2[0] = 0
 
    for i in range(N):
        dp1[i + 1] = max(dp1[i], dp2[i] + arr[i])
        dp2[i + 1] = max(dp2[i], dp1[i] - arr[i])
 
    # Return the maximum
    return max(dp1[len(dp1) - 1], dp2[len(dp2) - 1])
 
# Driver Code
arr = [2, 3, 6, 5, 4, 7, 8]
N = len(arr)
 
print(maxScore(arr, N))
 
# This code is contributed by Saurabh Jaiswal

C#




// C# code to implement the above approach
using System;
class GFG
{
 
    static double INF = 1E9;
 
    // Function to find the maximum score
    // achieved by alternative minus-plus
    // of elements of a subsequence in
    // the given array
    public static int maxScore(int []arr, int N) {
        int []dp1 = new int[N + 1];
        int []dp2 = new int[N + 1];
 
        dp1[0] = (int) -INF;
        dp2[0] = 0;
 
        for (int i = 0; i < N; ++i) {
            dp1[i + 1] = Math.Max(dp1[i], dp2[i] + arr[i]);
            dp2[i + 1] = Math.Max(dp2[i], dp1[i] - arr[i]);
        }
       
        // Return the maximum
        return Math.Max(dp1[dp1.Length - 1], dp2[dp2.Length - 1]);
    }
 
    // Driver Code
    public static void Main() {
        int []arr = { 2, 3, 6, 5, 4, 7, 8 };
        int N = arr.Length;
 
        Console.Write(maxScore(arr, N));
    }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript




<script>
 
        // JavaScript Program to implement
        // the above approach
        let INF = 1e9
 
        // Function to find the maximum score
        // achieved by alternative minus-plus
        // of elements of a subsequence in
        // the given array
        function maxScore(arr, N) {
            let dp1 = new Array(N + 1);
            let dp2 = new Array(N + 1);
 
            dp1[0] = -INF;
            dp2[0] = 0;
 
            for (let i = 0; i < N; ++i) {
                dp1[i + 1] = Math.max(dp1[i], dp2[i] + arr[i]);
                dp2[i + 1] = Math.max(dp2[i], dp1[i] - arr[i]);
            }
             
            // Return the maximum
            return Math.max(dp1[dp1.length - 1], dp2[dp2.length - 1]);
        }
 
        // Driver Code
        let arr = [2, 3, 6, 5, 4, 7, 8]
        let N = arr.length;
 
        document.write(maxScore(arr, N));
 
    // This code is contributed by Potta Lokesh
    </script>

 
 

Output
10

 

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

 


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