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

Narendra Kumar

Updated on: 31-Dec-2019

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