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Maximum sum of the array after dividing it into three segments
  • Last Updated : 07 Aug, 2019

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 <bits/stdc++.h>
using namespace std;
  
// Function to find maximum sum of array 
// after dividing it into three segments
int 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 reuired 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 code
int main()
{
    int a[] = { -6, 10, -3, 10, -2 };
  
    int n = sizeof(a) / sizeof(a[0]);
      
    // 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 segments
import java.util.*;
  
class GFG 
{
  
// Function to find maximum sum of array 
// after dividing it into three segments
static 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 reuired 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 code
public 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 segments
def 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 reuired 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 code
a = [-6, 10, -3, 10, -2]
  
n = len(a)
  
# Function call
print("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 segments
using System;
  
class GFG 
{
  
// Function to find maximum sum of array 
// after dividing it into three segments
static 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 reuired 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 code
public 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




<?php 
// PHP 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 segments 
function Max_Sum($a, $n
    // To store sum upto ith index 
    $b = array(); 
    $S = 0; 
    $res = 0; 
      
    // Traverse through the array 
    for ($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 reuired 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 = $n - 1; $i >= 0; --$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 call 
echo ("Maximum sum is: ");
echo Max_Sum($a, $n);
  
// This code is contributed by Naman_garg. 
?> 
Output:
Maximum sum is: 25

Time complexity : O(N)

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