Maximize the sum of array after multiplying a prefix and suffix by -1

Given an array arr[] of length N, the task is to maximize the sum of all the elements of the array by performing the following operations at most once. 

• Choose a prefix of the array and multiply all the elements by -1.
• Choose a suffix of the array and multiply all the elements by -1.

Examples:  

Input: arr[] = {-1, -2, -3} 
Output: 6 
Explanation: 
Operation 1: Prefix of array - {-1, -2, -3} 
can be multiplied with -1 to get the maximum sum. 
Array after Operation: {1, 2, 3} 
Sum = 1 + 2 + 3 = 6

Input: arr[] = {-4, 2, 0, 5, 0} 
Output: 11 
Explanation: 
Operation 1: Prefix of array - {-4} 
can be multiplied with -1 to get the maximum sum. 
Array after operation: {4, 2, 0, 5, 0} 
Sum = 4 + 2 + 0 + 5 + 0 = 11 

Approach: The key observation in the problem is if the chosen range of the prefix and suffix intersect, then the elements of intersection portion have same sign. Due to which it is always better to choose non-intersecting ranges of the prefix and suffix array. Below is the illustration of the steps:  

• It is easily been observed that there will be a portion/subarray in the array whose sum is the same as the original and the sum of the other elements is reversed. So the new sum of the array will be:

// X - Sum of subarray which is not in
//     the range of the prefix and suffix
// S - Sum of the original array

New Sum = X + -1*(S - X) = 2*X - S 

• Hence, the idea is to maximize the value of X to get the maximum sum because S is the constant value which cannot be changed. This can be achieved with the help of the Kadane's Algorithm.

Below is the implementation of the above approach: 

// C++ implementation to find the
// maximum sum of the array by 
// multiplying the prefix and suffix
// of the array by -1

#include <bits/stdc++.h>

using namespace std;

// Kadane's algorithm to find 
// the maximum subarray sum
int maxSubArraySum(int a[], int size)
{
    int max_so_far = INT_MIN, 
         max_ending_here = 0;
    
    // Loop to find the maximum subarray
    // array sum in the given array
    for (int i = 0; i < size; i++) {
        max_ending_here = 
             max_ending_here + a[i];
        if (max_ending_here < 0)
            max_ending_here = 0;
        if (max_so_far < max_ending_here)
            max_so_far = max_ending_here;
    }
    return max_so_far;
}

// Function to find the maximum
// sum of the array by multiplying
// the prefix and suffix by -1 
int maxSum(int a[], int n)
{
    
    // Total initial sum
    int S = 0;
    
    // Loop to find the maximum
    // sum of the array
    for (int i = 0; i < n; i++)
        S += a[i];
    int X = maxSubArraySum(a, n);
    
    // Maximum value
    return 2 * X - S;
}

// Driver Code
int main()
{
    int a[] = { -1, -2, -3 };
    int n = sizeof(a) / sizeof(a[0]);
    int max_sum = maxSum(a, n);
    cout << max_sum;
    return 0;
}

// Java implementation to find the
// maximum sum of the array by 
// multiplying the prefix and suffix
// of the array by -1
import java.io.*;
class GFG
{
     
    // Kadane's algorithm to find 
    // the maximum subarray sum
    static int maxSubArraySum(int a[], int size)
    {
        int max_so_far = Integer.MIN_VALUE, 
             max_ending_here = 0;
         
        // Loop to find the maximum subarray
        // array sum in the given array
        for (int i = 0; i < size; i++) {
            max_ending_here = 
                 max_ending_here + a[i];
            if (max_ending_here < 0)
                max_ending_here = 0;
            if (max_so_far < max_ending_here)
                max_so_far = max_ending_here;
        }
        return max_so_far;
    }
     
    // Function to find the maximum
    // sum of the array by multiplying
    // the prefix and suffix by -1 
    static int maxSum(int a[], int n)
    {
         
        // Total initial sum
        int S = 0;
        int i; 

        // Loop to find the maximum
        // sum of the array
        for (i = 0; i < n; i++)
            S += a[i];
        int X = maxSubArraySum(a, n);
         
        // Maximum value
        return 2 * X - S;
    }
     
    // Driver Code
    public static void main(String []args)
    {
        int a[] = { -1, -2, -3 };
        int n = a.length;
        int max_sum = maxSum(a, n);
        System.out.print(max_sum);
    }
}

// This code is contributed by chitranayal

# Python3 implementation to find the
# maximum sum of the array by
# multiplying the prefix and suffix
# of the array by -1

# Kadane's algorithm to find
# the maximum subarray sum
def maxSubArraySum(a, size):
    max_so_far = -10**9
    max_ending_here = 0

    # Loop to find the maximum subarray
    # array sum in the given array
    for i in range(size):
        max_ending_here = max_ending_here + a[i]
        if (max_ending_here < 0):
            max_ending_here = 0
        if (max_so_far < max_ending_here):
            max_so_far = max_ending_here
    return max_so_far

# Function to find the maximum
# sum of the array by multiplying
# the prefix and suffix by -1
def maxSum(a, n):

    # Total initial sum
    S = 0

    # Loop to find the maximum
    # sum of the array
    for i in range(n):
        S += a[i]
    X = maxSubArraySum(a, n)

    # Maximum value
    return 2 * X - S

# Driver Code
if __name__ == '__main__':
    a=[-1, -2, -3]
    n= len(a)
    max_sum = maxSum(a, n)
    print(max_sum)

# This code is contributed by mohit kumar 29

// C# implementation to find the
// maximum sum of the array by 
// multiplying the prefix and suffix
// of the array by -1
using System;

class GFG
{
      
    // Kadane's algorithm to find 
    // the maximum subarray sum
    static int maxSubArraySum(int []a, int size)
    {
        int max_so_far = int.MinValue, 
             max_ending_here = 0;
          
        // Loop to find the maximum subarray
        // array sum in the given array
        for (int i = 0; i < size; i++) {
            max_ending_here = 
                 max_ending_here + a[i];
            if (max_ending_here < 0)
                max_ending_here = 0;
            if (max_so_far < max_ending_here)
                max_so_far = max_ending_here;
        }
        return max_so_far;
    }
      
    // Function to find the maximum
    // sum of the array by multiplying
    // the prefix and suffix by -1 
    static int maxSum(int []a, int n)
    {
          
        // Total initial sum
        int S = 0;
        int i; 
 
        // Loop to find the maximum
        // sum of the array
        for (i = 0; i < n; i++)
            S += a[i];
        int X = maxSubArraySum(a, n);
          
        // Maximum value
        return 2 * X - S;
    }
      
    // Driver Code
    public static void Main(String []args)
    {
        int []a = { -1, -2, -3 };
        int n = a.Length;
        int max_sum = maxSum(a, n);
        Console.Write(max_sum);
    }
}

// This code is contributed by sapnasingh4991

<script>

// Javascript implementation to find the
// maximum sum of the array by 
// multiplying the prefix and suffix
// of the array by -1    

// Kadane's algorithm to find
// the maximum subarray sum
function maxSubArraySum(a, size)
{
    var max_so_far = Number.MIN_VALUE, 
        max_ending_here = 0;

    // Loop to find the maximum subarray
    // array sum in the given array
    for(i = 0; i < size; i++)
    {
        max_ending_here = max_ending_here + a[i];
        
        if (max_ending_here < 0)
            max_ending_here = 0;
            
        if (max_so_far < max_ending_here)
            max_so_far = max_ending_here;
    }
    return max_so_far;
}

// Function to find the maximum
// sum of the array by multiplying
// the prefix and suffix by -1
function maxSum(a, n) 
{
    
    // Total initial sum
    var S = 0;
    var i;

    // Loop to find the maximum
    // sum of the array
    for(i = 0; i < n; i++)
        S += a[i];
        
    var X = maxSubArraySum(a, n);

    // Maximum value
    return 2 * X - S;
}

// Driver Code
var a = [ -1, -2, -3 ];
var n = a.length;
var max_sum = maxSum(a, n);

document.write(max_sum);

// This code is contributed by aashish1995

</script>

6

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