Maximum sum of all elements of array after performing given operations

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: 

// CPP program to find the maximum
// sum after given operations

#include <bits/stdc++.h>
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[0]);

    cout << maxSum(a, n);

    return 0;
}

// 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 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# 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
// PHP program to find the maximum
// sum after given operations

// Function to calculate Maximum Subarray 
// Sum or Kadane's Algorithm
function  maxSubArraySum($a, $size)
{
    $max_so_far = PHP_INT_MIN;
    $max_ending_here = 0;

    for ($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
function 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 = 0; $i < $n; $i++)
        $S += $a[$i];

    return (2 * $S1 - $S);
}

// Driver Code
$a = array(-35, 32, -24, 0, 
            27, -10, 0, -19);

// size of an array
$n = sizeof($a);

echo( maxSum($a, $n));

// This code is contributed 
// by Mukul Singh

<script>
// javascript program to find the maximum
// sum after given operations    

// Function to calculate Maximum Subarray Sum
    // or Kadane's Algorithm
    function maxSubArraySum(a , size)
    {
        var max_so_far = Number.MIN_VALUE, max_ending_here = 0;

        for (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
    function maxSum(a, n) 
    {
    
        // To store sum of all elements
        var S = 0;

        // Maximum sum of a subarray
        var S1 = maxSubArraySum(a, n);

        // Calculate the sum of all elements
        for (i = 0; i < n; i++)
            S += a[i];

        return (2 * S1 - S);
    }

    // Driver Code
        var a = [ -35, 32, -24, 0, 27, -10, 0, -19 ];

        // size of an array
        var n = a.length;

        document.write(maxSum(a, n));

// This code is contributed by todaysgaurav. 
</script>

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.