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# Maximum sum of all elements of array after performing given operations

• Difficulty Level : Medium
• Last Updated : 12 Sep, 2022

Given an array of integers. The task is to find the maximum sum of all the elements of the array after performing the given two operations once each.
The operations are:

1. Select some(possibly none) continuous elements from the beginning of the array and multiply by -1.
2. Select some(possibly none) continuous elements from the end of the array and multiply by -1.

Examples:

```Input : arr[] = {-1, 10, -5, 10, -2}
Output : 18
After 1st operation : 1 10 -5 10 -2
After 2nd operation : 1 10 -5 10 2

Input : arr[] = {-9, -8, -7}
Output : 24
After 1st operation : 9 8 -7
After 2nd operation : 9 8 7```

Approach: This problem can be solved in linear time, using the following idea:

• Let the sum of elements A1 .. An be equal to S. Then when inverting signs we get -A1, -A2 .. -An, and the sum is thereafter changed to -S, i.e. sum of elements on the segment will just change its’ sign when inverting the whole segment’s signs.
• Consider the initial problem as follows: choose a consecutive subsequence, and invert all the numbers remaining out of it.
• Find the Maximum subarray sum using Kadane’ Algorithm.
• Keep that subarray intact and multiply the rest with -1.
• Considering the sum of the whole array as S, and the largest sum contiguous subarray as S1, the total sum will be equal to -(S-S1) + S1 = 2*S1 – S. This is the required sum.

Below is the implementation of the above approach:

## C++

 `// CPP program to find the maximum``// sum after given operations` `#include ``using` `namespace` `std;` `// Function to calculate Maximum Subarray Sum``// or Kadane's Algorithm``int` `maxSubArraySum(``int` `a[], ``int` `size)``{``    ``int` `max_so_far = INT_MIN, max_ending_here = 0;` `    ``for` `(``int` `i = 0; i < size; i++) {``        ``max_ending_here = max_ending_here + a[i];``        ``if` `(max_so_far < max_ending_here)``            ``max_so_far = max_ending_here;` `        ``if` `(max_ending_here < 0)``            ``max_ending_here = 0;``    ``}``    ``return` `max_so_far;``}` `// Function to find the maximum``// sum after given operations``int` `maxSum(``int` `a[], ``int` `n)``{``    ``// To store sum of all elements``    ``int` `S = 0;` `    ``// Maximum sum of a subarray``    ``int` `S1 = maxSubArraySum(a, n);` `    ``// Calculate the sum of all elements``    ``for` `(``int` `i = 0; i < n; i++)``        ``S += a[i];` `    ``return` `(2 * S1 - S);``}` `// Driver Code``int` `main()``{``    ``int` `a[] = { -35, 32, -24, 0, 27, -10, 0, -19 };` `    ``// size of an array``    ``int` `n = ``sizeof``(a) / ``sizeof``(a);` `    ``cout << maxSum(a, n);` `    ``return` `0;``}`

## Java

 `// Java program to find the maximum``// sum after given operations` `import` `java.io.*;` `class` `GFG {``  ` `// Function to calculate Maximum Subarray Sum``// or Kadane's Algorithm``static` `int` `maxSubArraySum(``int` `a[], ``int` `size)``{``    ``int` `max_so_far = Integer.MIN_VALUE, max_ending_here = ``0``;` `    ``for` `(``int` `i = ``0``; i < size; i++) {``        ``max_ending_here = max_ending_here + a[i];``        ``if` `(max_so_far < max_ending_here)``            ``max_so_far = max_ending_here;` `        ``if` `(max_ending_here < ``0``)``            ``max_ending_here = ``0``;``    ``}``    ``return` `max_so_far;``}` `// Function to find the maximum``// sum after given operations``static` `int` `maxSum(``int` `a[], ``int` `n)``{``    ``// To store sum of all elements``    ``int` `S = ``0``;` `    ``// Maximum sum of a subarray``    ``int` `S1 = maxSubArraySum(a, n);` `    ``// Calculate the sum of all elements``    ``for` `(``int` `i = ``0``; i < n; i++)``        ``S += a[i];` `    ``return` `(``2` `* S1 - S);``}` `// Driver Code`  `    ``public` `static` `void` `main (String[] args) {``    ``int` `a[] = { -``35``, ``32``, -``24``, ``0``, ``27``, -``10``, ``0``, -``19` `};` `    ``// size of an array``    ``int` `n = a.length;` `    ``System.out.println( maxSum(a, n));``    ``}``}``// This code is contributed by inder_verma`

## Python3

 `# Python3 program to find the maximum``# sum after given operations``import` `sys` `# Function to calculate Maximum``# Subarray Sum or Kadane's Algorithm``def` `maxSubArraySum(a, size) :``        ` `    ``max_so_far ``=` `-``(sys.maxsize ``-` `1``)``    ``max_ending_here ``=` `0` `    ``for` `i ``in` `range``(size) :``        ` `        ``max_ending_here ``=` `max_ending_here ``+` `a[i]``        ` `        ``if` `(max_so_far < max_ending_here) :``                ``max_so_far ``=` `max_ending_here` `        ``if` `(max_ending_here < ``0``) :``                ``max_ending_here ``=` `0``    ` `    ``return` `max_so_far` `# Function to find the maximum``# sum after given operations``def` `maxSum(a, n) :``    ` `    ``# To store sum of all elements``    ``S ``=` `0``;` `    ``# Maximum sum of a subarray``    ``S1 ``=` `maxSubArraySum(a, n)` `    ``# Calculate the sum of all elements``    ``for` `i ``in` `range``(n) :``        ``S ``+``=` `a[i]` `    ``return` `(``2` `*` `S1 ``-` `S)` `# Driver Code``if` `__name__ ``=``=` `"__main__"` `:` `    ``a ``=` `[ ``-``35``, ``32``, ``-``24``, ``0``,``           ``27``, ``-``10``, ``0``, ``-``19` `]` `    ``# size of an array``    ``n ``=` `len``(a)` `    ``print``(maxSum(a, n))` `# This code is contributed by Ryuga`

## C#

 `// C# program to find the maximum``// sum after given operations` `using` `System;` `class` `GFG {` `// Function to calculate Maximum Subarray Sum``// or Kadane's Algorithm``static` `int` `maxSubArraySum(``int` `[]a, ``int` `size)``{``    ``int` `max_so_far = ``int``.MinValue, max_ending_here = 0;` `    ``for` `(``int` `i = 0; i < size; i++) {``        ``max_ending_here = max_ending_here + a[i];``        ``if` `(max_so_far < max_ending_here)``            ``max_so_far = max_ending_here;` `        ``if` `(max_ending_here < 0)``            ``max_ending_here = 0;``    ``}``    ``return` `max_so_far;``}` `// Function to find the maximum``// sum after given operations``static` `int` `maxSum(``int` `[]a, ``int` `n)``{``    ``// To store sum of all elements``    ``int` `S = 0;` `    ``// Maximum sum of a subarray``    ``int` `S1 = maxSubArraySum(a, n);` `    ``// Calculate the sum of all elements``    ``for` `(``int` `i = 0; i < n; i++)``        ``S += a[i];` `    ``return` `(2 * S1 - S);``}` `// Driver Code`  `    ``public` `static` `void` `Main () {``    ``int` `[]a = { -35, 32, -24, 0, 27, -10, 0, -19 };` `    ``// size of an array``    ``int` `n = a.Length;` `    ``Console.WriteLine( maxSum(a, n));``    ``}``}``// This code is contributed by inder_verma`

## PHP

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## Javascript

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Output

`99`

Complexity Analysis:

• Time Complexity: O(n), where n represents the size of the given array.
• Auxiliary Space: O(1), no extra space is required, so it is a constant.

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