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 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[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 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
<?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 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 = $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. ?> |
Javascript
<script> // Javascript 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 let b = new Array(n); let S = 0; let res = 0; // Traverse through the array for (let 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 let 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) let g = 0; // For third segment for (let i = n - 1; i >= 0; --i) { g -= a[i]; ans = Math.max(ans, g + b[i]); } // return the required answer return ans; } let a = [ -6, 10, -3, 10, -2 ]; let n = a.length; // Function call document.write("Maximum sum is: " + Max_Sum(a, n));</script> |
Output:
Maximum sum is: 25
Complexity Analysis:
Time complexity: O(N), as we are having 2 passes of the loop of size N inside the Max_Sum() function, so time complexity will be O(N+N)->O(N).
Space Complexity: O(N), as we have created an array of size N inside the Max_Sum() function.
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