Skip to content
Related Articles
Maximum possible array sum after performing the given operation
• Last Updated : 05 May, 2021

Given an array arr[] of size N, the task is to find the maximum sum of the elements of the array after applying the given operation any number of times. In a single operation, choose an index 1 ≤ i < N and multiply both arr[i] and arr[i – 1] by -1.
Examples:

Input: arr[] = {-10, 5, -4}
Output: 19
Perform the operation for i = 1 and
the array becomes {10, -5, -4}
Perform the operation for i = 2 and
the array becomes {10, 5, 4}
10 + 5 + 4 = 19
Input: arr[] = {10, -4, -8, -11, 3}
Output: 30

Approach: This problem can be solved using dynamic programming. Since it is useless to choose and flip the same arr[i], we will consider flipping at most once for each element of the array in order from the left. Let dp[i] represents the maximum possible sum up to the ith index without flipping the ith index. dp[i] represents the maximum possible sum up to the ith index with flipping the ith index. So, dp(n, 0) is our required answer.
Base conditions:

1. dp = 0
2. dp = INT_MIN

Recurrence relation:

1. dp[i + 1] = max(dp[i] + arr[i], dp[i] – arr[i])
2. dp[i + 1] = max(dp[i] – arr[i], dp[i] + arr[i])

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to return the maximum possible``// sum after performing the given operation``int` `max_sum(``int` `a[], ``int` `n)``{``    ``// Dp vector to store the answer``    ``vector > dp(n + 1,``                            ``vector<``int``>(2, 0));` `    ``// Base value``    ``dp = 0, dp = -999999;` `    ``for` `(``int` `i = 0; i <= n - 1; i++) {``        ``dp[i + 1] = max(dp[i] + a[i],``                           ``dp[i] - a[i]);``        ``dp[i + 1] = max(dp[i] - a[i],``                           ``dp[i] + a[i]);``    ``}` `    ``// Return the maximum sum``    ``return` `dp[n];``}` `// Driver code``int` `main()``{``    ``int` `a[] = { -10, 5, -4 };``    ``int` `n = ``sizeof``(a) / ``sizeof``(a);` `    ``cout << max_sum(a, n);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``class` `GFG``{` `// Function to return the maximum possible``// sum after performing the given operation``static` `int` `max_sum(``int` `a[], ``int` `n)``{``    ``// Dp vector to store the answer``    ``int` `[][]dp = ``new` `int``[n + ``1``][``2``];``                        ` `    ``// Base value``    ``dp[``0``][``0``] = ``0``; dp[``0``][``1``] = -``999999``;` `    ``for` `(``int` `i = ``0``; i <= n - ``1``; i++)``    ``{``        ``dp[i + ``1``][``0``] = Math.max(dp[i][``0``] + a[i],``                                ``dp[i][``1``] - a[i]);``        ``dp[i + ``1``][``1``] = Math.max(dp[i][``0``] - a[i],``                                ``dp[i][``1``] + a[i]);``    ``}` `    ``// Return the maximum sum``    ``return` `dp[n][``0``];``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `a[] = { -``10``, ``5``, -``4` `};``    ``int` `n = a.length;` `    ``System.out.println(max_sum(a, n));``}``}` `// This code is contributed by 29AjayKumar`

## Python3

 `# Python implementation of the approach` `# Function to return the maximum possible``# sum after performing the given operation``def` `max_sum(a, n):``    ``# Dp vector to store the answer``    ``dp ``=` `[[``0` `for` `i ``in` `range``(``2``)] ``for` `j ``in` `range``(n``+``1``)]``                        ` `    ``# Base value``    ``dp[``0``][``0``] ``=` `0``; dp[``0``][``1``] ``=` `-``999999``;` `    ``for` `i ``in` `range``(``0``, n):``        ``dp[i ``+` `1``][``0``] ``=` `max``(dp[i][``0``] ``+` `a[i],``                                ``dp[i][``1``] ``-` `a[i]);``        ``dp[i ``+` `1``][``1``] ``=` `max``(dp[i][``0``] ``-` `a[i],``                                ``dp[i][``1``] ``+` `a[i]);` `    ``# Return the maximum sum``    ``return` `dp[n][``0``];` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ``a ``=` `[``-``10``, ``5``, ``-``4` `];``    ``n ``=` `len``(a);` `    ``print``(max_sum(a, n));``    ` `# This code is contributed by 29AjayKumar`

## C#

 `// C# implementation of the approach``using` `System;``    ` `class` `GFG``{` `// Function to return the maximum possible``// sum after performing the given operation``static` `int` `max_sum(``int` `[]a, ``int` `n)``{``    ``// Dp vector to store the answer``    ``int` `[,]dp = ``new` `int``[n + 1, 2];``                        ` `    ``// Base value``    ``dp[0, 0] = 0; dp[0, 1] = -999999;` `    ``for` `(``int` `i = 0; i <= n - 1; i++)``    ``{``        ``dp[i + 1, 0] = Math.Max(dp[i, 0] + a[i],``                                ``dp[i, 1] - a[i]);``        ``dp[i + 1, 1] = Math.Max(dp[i, 0] - a[i],``                                ``dp[i, 1] + a[i]);``    ``}` `    ``// Return the maximum sum``    ``return` `dp[n, 0];``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``int` `[]a = { -10, 5, -4 };``    ``int` `n = a.Length;` `    ``Console.WriteLine(max_sum(a, n));``}``}` `// This code is contributed by PrinciRaj1992`

## Javascript

 ``
Output:
`19`

Time Complexity: O(N)

Auxiliary Space: O(N)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with industry experts, please refer Geeks Classes Live

My Personal Notes arrow_drop_up