# Maximizing array sum with given operation in C++

## Description

There is an array of (2 * n – 1) integers. We can change sign of exactly n elements in the array. In other words, we can select exactly n array elements, and multiply each of them by -1. Find the maximum sum of the array.

## Example

If input array is {-2, 100, -3} then we can obtain maximum changing sign of -2 and -3. After changing sign array becomes −

{2, 100, 3} and maximum sum of this array is 105.

## Algorithm

• Count negative numbers
• Calculate the sum of the array by taking absolute values of the numbers.
• Find the minimum number of the array by taking absolute values of the numbers
• Check if the no. of negative numbers is odd and the value of n is even then subtracting two times m from the sum and this will be maximum sum of the array else, the value of sum will be the maximum sum of the array
• Repeat above steps (2 * n – 1) times

## Example

Let us now see an example −

Live Demo

#include <bits/stdc++.h>
using namespace std;
int getMaxSum(int *arr, int n) {
int negtiveCnt = 0;
int sum = 0;
int m = INT_MAX;
for (int i = 0; i < 2 * n - 1; ++i) {
if (arr[i] < 0) {
++negtiveCnt;
}
sum = sum + abs(arr[i]);
m = min(m, abs(arr[i]));
}
if (negtiveCnt % 2 && n % 2 == 0) {
sum = sum - 2 * m;
return sum;
}
return sum;
}
int main() {
int arr[] = {-2, 100, -3};
int n = 2;
cout << "Maximum sum = " << getMaxSum(arr, n) << endl;
return 0;
}

## Output

Maximum sum = 105

Updated on: 31-Dec-2019

70 Views 