Maximize sum of an Array by flipping sign of all elements of a single subarray

Given an array arr[] of N integers, the task is to find the maximum sum of the array that can be obtained by flipping signs of any subarray of the given array at most once.

Examples:

Input: arr[] = {-2, 3, -1, -4, -2} 
Output: 8
Explanation: 
Flipping the signs of subarray {-1, -4, -2} modifies the array to {-2, 3, 1, 4, 2}. Therefore, the sum of the array = -2 + 3 + 1 + 4 + 2 = 8, which is the maximum possible.

Input: arr[] = {1, 2, -10, 2, -20}
Output: 31 
Explanation: 
Flipping the signs of subarray {-10, 2, -20} modifies the array to {1, 2, 10, -2, 20}. Therefore, the sum of the array = 1 + 2 + 10 – 2 + 20 = 31, which is the maximum possible.

Naive Approach: The simplest approach is to calculate the total sum of the array and then generate all possible subarrays. Now, for each subarray {A[i], … A[j]}, subtract its sum, sum(A[i], …, A[j]), from the total sum and flip the signs of the subarray elements. After flipping the subarray, add the sum of the flipped subarray, i.e. (-1 * sum(A[i], …, A[j])), to the total sum. Below are the steps:



  1. Find the total sum of the original array (say total_sum) and store it.
  2. Now, for all possible subarrays find the maximum of total_sum – 2 * sum(i, j).

Below is the implementation of the above approach:

C++

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// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the maximum sum
// after flipping a subarray
int maxSumFlip(int a[], int n)
{
     
    // Stores the total sum of array
    int total_sum = 0;
    for(int i = 0; i < n; i++)
        total_sum += a[i];
 
    // Initialize the maximum sum
    int max_sum = INT_MIN;
 
    // Iterate over all possible subarrays
    for(int i = 0; i < n; i++)
    {
         
        // Initialize sum of the subarray
        // before flipping sign
        int sum = 0;
 
        // Initialize sum of subarray
        // after flipping sign
        int flip_sum = 0;
 
        for(int j = i; j < n; j++)
        {
 
            // Calculate the sum of
            // original subarray
            sum += a[j];
 
            // Subtract the original
            // subarray sum and add
            // the flipped subarray
            // sum to the total sum
            max_sum = max(max_sum,
                        total_sum - 2 * sum);
        }
    }
 
    // Return the max_sum
    return max(max_sum, total_sum);
}
 
// Driver Code
int main()
{
    int arr[] = { -2, 3, -1, -4, -2 };
    int N = sizeof(arr) / sizeof(int);
     
    cout << maxSumFlip(arr, N);
}
 
// This code is contributed by sanjoy_62

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Java

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// Java program for the above approach
 
import java.io.*;
 
public class GFG {
 
    // Function to find the maximum sum
    // after flipping a subarray
    public static int
    maxSumFlip(int a[], int n)
    {
        // Stores the total sum of array
        int total_sum = 0;
        for (int i = 0; i < n; i++)
            total_sum += a[i];
 
        // Initialize the maximum sum
        int max_sum = Integer.MIN_VALUE;
 
        // Iterate over all possible subarrays
        for (int i = 0; i < n; i++) {
 
            // Initialize sum of the subarray
            // before flipping sign
            int sum = 0;
 
            // Initialize sum of subarray
            // after flipping sign
            int flip_sum = 0;
 
            for (int j = i; j < n; j++) {
 
                // Calculate the sum of
                // original subarray
                sum += a[j];
 
                // Subtract the original
                // subarray sum and add
                // the flipped subarray
                // sum to the total sum
                max_sum = Math.max(max_sum,
                                   total_sum - 2 * sum);
            }
        }
 
        // Return the max_sum
        return Math.max(max_sum, total_sum);
    }
 
    // Driver Code
    public static void
    main(String args[])
    {
        int arr[] = { -2, 3, -1, -4, -2 };
        int N = arr.length;
        System.out.println(maxSumFlip(arr, N));
    }
}

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Python3

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# Python3 program for the above approach
import sys
 
# Function to find the maximum sum
# after flipping a subarray
def maxSumFlip(a, n):
     
    # Stores the total sum of array
    total_sum = 0
    for i in range(n):
        total_sum += a[i]
 
    # Initialize the maximum sum
    max_sum = -sys.maxsize - 1
 
    # Iterate over all possible subarrays
    for i in range(n):
 
        # Initialize sum of the subarray
        # before flipping sign
        sum = 0
 
        # Initialize sum of subarray
        # after flipping sign
        flip_sum = 0
 
        for j in range(i, n):
 
            # Calculate the sum of
            # original subarray
            sum += a[j]
 
            # Subtract the original
            # subarray sum and add
            # the flipped subarray
            # sum to the total sum
            max_sum = max(max_sum,
                        total_sum - 2 * sum)
         
    # Return the max_sum
    return max(max_sum, total_sum)
 
# Driver Code
arr = [ -2, 3, -1, -4, -2 ]
N = len(arr)
 
print(maxSumFlip(arr, N))
 
# This code is contributed by sanjoy_62

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C#

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// C# program for the above approach
using System;
class GFG{
 
// Function to find the maximum sum
// after flipping a subarray
public static int maxSumFlip(int []a, int n)
{
     
    // Stores the total sum of array
    int total_sum = 0;
    for(int i = 0; i < n; i++)
        total_sum += a[i];
 
    // Initialize the maximum sum
    int max_sum = int.MinValue;
 
    // Iterate over all possible subarrays
    for(int i = 0; i < n; i++)
    {
         
        // Initialize sum of the subarray
        // before flipping sign
        int sum = 0;
 
        // Initialize sum of subarray
        // after flipping sign
        //int flip_sum = 0;
 
        for(int j = i; j < n; j++)
        {
             
            // Calculate the sum of
            // original subarray
            sum += a[j];
 
            // Subtract the original
            // subarray sum and add
            // the flipped subarray
            // sum to the total sum
            max_sum = Math.Max(max_sum,
                               total_sum - 2 * sum);
        }
    }
 
    // Return the max_sum
    return Math.Max(max_sum, total_sum);
}
 
// Driver Code
public static void Main(String []args)
{
    int []arr = { -2, 3, -1, -4, -2 };
    int N = arr.Length;
     
    Console.WriteLine(maxSumFlip(arr, N));
}
}
 
// This code is contributed by 29AjayKumar

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Output: 

8

Time Complexity: O(N2
Auxiliary Space: O(1)

Efficient Approach: From the above approach, it can eb observed that, to obtain maximum array sum, (2 * subarray sum) needs to be maximized for all subarrays. This can be done by using Dynamic Programming. Below are the steps:

  1. Create a new array brr[], where brr[i] = (-1 * arr[i]) – arr[i]
  2. Find the maximum sum subarray from brr[] using Kadane’s Algorithm
  3. This maximizes the contribution of (2 * sum) over all subarrays.
  4. Add the maximum contribution to the total sum of the array.

Below is the implementation of the above approach:

C++

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// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the maximum sum
// after flipping a subarray
int maxSumFlip(int a[], int n)
{
     
    // Find the total sum of array
    int total_sum = 0;
     
    for(int i = 0; i < n; i++)
        total_sum += a[i];
 
    // Using Kadane's Algorithm
    int max_ending_here = -a[0] - a[0];
    int curr_sum = -a[0] - a[0];
 
    for(int i = 1; i < n; i++)
    {
         
        // Either extend previous
        // sub_array or start
        // new subarray
        curr_sum = max(curr_sum + (-a[i] - a[i]),
                                  (-a[i] - a[i]));
 
        // Keep track of max_sum array
        max_ending_here = max(max_ending_here,
                              curr_sum);
    }
 
    // Add the sum to the total_sum
    int max_sum = total_sum +
                  max_ending_here;
 
    // Check max_sum was maximum
    // with flip or without flip
    max_sum = max(max_sum, total_sum);
 
    // Return max_sum
    return max_sum;
}
 
// Driver code  
int main()
{
    int arr[] = { -2, 3, -1, -4, -2 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function call
    cout << maxSumFlip(arr, N) << endl;
 
    return 0;
}
 
// This code is contributed by divyeshrabadiya07

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Java

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// Java program for the above approach
 
import java.util.*;
public class Main {
 
    // Function to find the maximum sum
    // after flipping a subarray
    public static int maxSumFlip(int a[],
                                 int n)
    {
        // Find the total sum of array
        int total_sum = 0;
        for (int i = 0; i < n; i++)
            total_sum += a[i];
 
        // Using Kadane's Algorithm
        int max_ending_here = -a[0] - a[0];
        int curr_sum = -a[0] - a[0];
 
        for (int i = 1; i < n; i++) {
 
            // Either extend previous
            // sub_array or start
            // new subarray
            curr_sum
                = Math.max(curr_sum + (-a[i] - a[i]),
                           (-a[i] - a[i]));
 
            // Keep track of max_sum array
            max_ending_here
                = Math.max(max_ending_here,
                           curr_sum);
        }
 
        // Add the sum to the total_sum
        int max_sum = total_sum
                      + max_ending_here;
 
        // Check max_sum was maximum
        // with flip or without flip
        max_sum = Math.max(max_sum, total_sum);
 
        // Return max_sum
        return max_sum;
    }
 
    // Driver Code
    public static void main(String args[])
    {
        int arr[] = { -2, 3, -1, -4, -2 };
        int N = a.length;
 
        // Function Call
        System.out.println(maxSumFlip(arr, N));
    }
}

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C#

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// C# program for the above approach
using System;
 
class GFG{
 
// Function to find the maximum sum
// after flipping a subarray
public static int maxSumFlip(int []a,
                             int n)
{
     
    // Find the total sum of array
    int total_sum = 0;
    for(int i = 0; i < n; i++)
        total_sum += a[i];
 
    // Using Kadane's Algorithm
    int max_ending_here = -a[0] - a[0];
    int curr_sum = -a[0] - a[0];
 
    for(int i = 1; i < n; i++)
    {
 
        // Either extend previous
        // sub_array or start
        // new subarray
        curr_sum = Math.Max(curr_sum +
                        (-a[i] - a[i]),
                        (-a[i] - a[i]));
 
        // Keep track of max_sum array
        max_ending_here = Math.Max(
                          max_ending_here,
                          curr_sum);
    }
 
    // Add the sum to the total_sum
    int max_sum = total_sum +
                  max_ending_here;
 
    // Check max_sum was maximum
    // with flip or without flip
    max_sum = Math.Max(max_sum,
                       total_sum);
 
    // Return max_sum
    return max_sum;
}
 
// Driver Code
public static void Main(String []args)
{
    int []arr = { -2, 3, -1, -4, -2 };
    int N = arr.Length;
 
    // Function call
    Console.WriteLine(maxSumFlip(arr, N));
}
}
 
// This code is contributed by 29AjayKumar

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Output: 

8

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

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