Maximum sum of the array after dividing it into three segments

• Last Updated : 06 Jul, 2021

Given an array a of size N. The task is to find the maximum sum of array possible by dividing the array into three segments such that each element in the first segment is multiplied by -1 and each element in the second segment is multiplied by 1 and each element in the third segment is multiplied by -1. The segments can intersect and any segment may include zero in it.

Examples:

Input : a[] = {-6, 10, -3, 10, -2}
Output : 25
Divide the segments as {-6}, {10, -3, 10}, {-2)

Input : a[] = {-6, -10}
Output : 16

Approach:
First we need is to calculate for all possible situations for every ith element where the division should be made.

• In the first traversal find if the ith element produces maximum sum by multiplying with -1 or keeping it as it is.
• Store all values in the array b.
• In the second traversal find maximum sum by decreasing a[i] and adding b[i] to it.

Below is the implementation of the above approach :

C++

 // C++ program to find maximum sum of array// after dividing it into three segments#include using namespace std; // Function to find maximum sum of array// after dividing it into three segmentsint Max_Sum(int a[], int n){    // To store sum upto ith index    int b[n];    int S = 0;    int res = 0;         // Traverse through the array    for (int i = 0; i < n; i++)    {        b[i] = res;        res += a[i];        S += a[i];                 // Get the maximum possible sum        res = max(res, -S);    }         // Store in the required answer    int ans = S;         // Maximum sum starting from left segment    // by choosing between keeping array element as    // it is or subtracting it    ans = max(ans, res);      // Finding maximum sum by decreasing a[i] and    // adding b[i] to it that means max(multiplying    // it by -1 or using b[i] value)    int g = 0;         // For third segment    for (int i = n - 1; i >= 0; --i) {        g -= a[i];        ans = max(ans, g + b[i]);    }         // return the required answer    return ans;} // Driver codeint main(){    int a[] = { -6, 10, -3, 10, -2 };     int n = sizeof(a) / sizeof(a);         // Function call    cout << "Maximum sum is: " << Max_Sum(a, n);     return 0;}

Java

 // Java program to find maximum sum of array// after dividing it into three segmentsimport java.util.*; class GFG{ // Function to find maximum sum of array// after dividing it into three segmentsstatic int Max_Sum(int a[], int n){    // To store sum upto ith index    int []b = new int[n];    int S = 0;    int res = 0;         // Traverse through the array    for (int i = 0; i < n; i++)    {        b[i] = res;        res += a[i];        S += a[i];                 // Get the maximum possible sum        res = Math.max(res, -S);    }         // Store in the required answer    int ans = S;         // Maximum sum starting from left segment    // by choosing between keeping array element as    // it is or subtracting it    ans = Math.max(ans, res);     // Finding maximum sum by decreasing a[i] and    // adding b[i] to it that means max(multiplying    // it by -1 or using b[i] value)    int g = 0;         // For third segment    for (int i = n - 1; i >= 0; --i)    {        g -= a[i];        ans = Math.max(ans, g + b[i]);    }         // return the required answer    return ans;} // Driver codepublic static void main(String[] args){    int a[] = { -6, 10, -3, 10, -2 };     int n = a.length;         // Function call    System.out.println("Maximum sum is: " +                            Max_Sum(a, n));}} // This code is contributed by Princi Singh

Python3

 # Python3 program to find# maximum sum of array after# dividing it into three segments # Function to find maximum sum of array# after dividing it into three segmentsdef Max_Sum(a, n):         # To store sum upto ith index    b = [0 for i in range(n)]    S = 0    res = 0     # Traverse through the array    for i in range(n):        b[i] = res        res += a[i]        S += a[i]         # Get the maximum possible sum        res = max(res, -S)     # Store in the required answer    ans = S     # Maximum sum starting from    # left segment by choosing between    # keeping array element as it is    # or subtracting it    ans = max(ans, res)     # Finding maximum sum by decreasing    # a[i] and adding b[i] to it    # that means max(multiplying it    # by -1 or using b[i] value)    g = 0     # For third segment    for i in range(n - 1, -1, -1):        g -= a[i]        ans = max(ans, g + b[i])     # return the required answer    return ans # Driver codea = [-6, 10, -3, 10, -2] n = len(a) # Function callprint("Maximum sum is:",          Max_Sum(a, n))            # This code is contributed# by Mohit Kumar

C#

 // C#+ program to find maximum sum of array// after dividing it into three segmentsusing System; class GFG{ // Function to find maximum sum of array// after dividing it into three segmentsstatic int Max_Sum(int []a, int n){    // To store sum upto ith index    int []b = new int[n];    int S = 0;    int res = 0;         // Traverse through the array    for (int i = 0; i < n; i++)    {        b[i] = res;        res += a[i];        S += a[i];                 // Get the maximum possible sum        res = Math.Max(res, -S);    }         // Store in the required answer    int ans = S;         // Maximum sum starting from left segment    // by choosing between keeping array element    // as it is or subtracting it    ans = Math.Max(ans, res);     // Finding maximum sum by decreasing a[i] and    // adding b[i] to it that means max(multiplying    // it by -1 or using b[i] value)    int g = 0;         // For third segment    for (int i = n - 1; i >= 0; --i)    {        g -= a[i];        ans = Math.Max(ans, g + b[i]);    }         // return the required answer    return ans;} // Driver codepublic static void Main(){    int []a = { -6, 10, -3, 10, -2 };     int n = a.Length;         // Function call    Console.WriteLine("Maximum sum is: " +                           Max_Sum(a, n));}} // This code is contributed by anuj_67..

PHP

 = 0; --\$i)    {        \$g -= \$a[\$i];        \$ans = max(\$ans, \$g + \$b[\$i]);    }         // return the required answer    return \$ans;} // Driver code\$a = array(-6, 10, -3, 10, -2 ); \$n = count(\$a); // Function callecho ("Maximum sum is: ");echo Max_Sum(\$a, \$n); // This code is contributed by Naman_garg.?>

Javascript


Output:
Maximum sum is: 25

Time complexity : O(N)

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