Maximum Matrix Sum

class Solution {
public:
    long long maximumSumOfMatrix(vector<vector<int>>& grid) {
        long long sumOfElements = 0;
        int smallestValue = INT_MAX;
        int numOfNegatives = 0;

        for (auto& row : grid) {
            for (int element : row) {
                sumOfElements += abs(element);
                if (element < 0) {
                    numOfNegatives++;
                }
                smallestValue = min(smallestValue, abs(element));
            }
        }

        // Compensate if odd negatives
        if (numOfNegatives % 2 != 0) {
            sumOfElements -= 2 * smallestValue;
        }

        return sumOfElements;
    }
};

The problem titled "Maximum Matrix Sum" is efficiently solved using C++. The key objective is to compute the maximum sum of a matrix after potentially flipping signs of some elements to maximize the total sum.

• The solution class has a function named maximumSumOfMatrix which takes a 2D vector grid representing the matrix.
• The function initializes sumOfElements to accumulate the total maximum sum possible, smallestValue to track the smallest absolute value encountered for possible adjustments, and numOfNegatives to count how many elements in the matrix are negative.
• It iterates through each row of the matrix, and for each element, it:
  • Adds the absolute value of the element to sumOfElements.
  • Counts the element as negative if it is less than zero.
  • Updates smallestValue with the minimum between the current smallestValue and the absolute value of the current element.
• Post the matrix traversal, the solution checks if the count of negative numbers is odd:
  • If it is, it compensates by subtracting twice the smallestValue from sumOfElements. This is based on the principle that flipping the smallest absolute value when having an odd number of negative values will yield a greater sum.
• Finally, it returns the sumOfElements as the maximum possible sum of the matrix.

This approach effectively leverages mathematical principles to compute the result in a straightforward but optimal manner without changing the actual matrix. It requires careful management of assignments and checks within loops to maintain time efficiency. This solution is robust and manages edge cases, such as matrices containing zero or all negative numbers.

class Solution {

    public long maxSumMatrix(int[][] data) {
        long sumTotal = 0;
        int minValue = Integer.MAX_VALUE;
        int countNegatives = 0;

        for (int[] rowData : data) {
            for (int cellVal : rowData) {
                sumTotal += Math.abs(cellVal);
                if (cellVal < 0) {
                    countNegatives++;
                }
                minValue = Math.min(minValue, Math.abs(cellVal));
            }
        }

        // Change sum based on odd count of negatives
        if (countNegatives % 2 != 0) {
            sumTotal -= 2 * minValue;
        }

        return sumTotal;
    }
}

The problem tackled here involves calculating the maximum sum possible for a matrix where the values can either be flipped to their negative counterparts or kept positive. The solution is implemented in Java and involves strategic manipulation of matrix values to achieve the highest possible sum.

• Begin by initializing sumTotal to track the aggregated absolute values of the matrix elements.

• Initialize minValue to the maximum possible integer value to facilitate finding the smallest absolute value in the matrix.

• countNegatives keeps track of how many negative numbers there are in the entire matrix.

• Iterate through each element of the matrix:

  • Add the absolute value of the current element to sumTotal.
  • If the element is negative, increment countNegatives.
  • Continuously update minValue with the smaller value between the current minValue and the absolute value of the current matrix element.

• After iterating through all elements:

  • Check if countNegatives is odd. If it is, subtract twice the minValue from sumTotal. This adjustment accounts for the possibility of one unpaired negative value that could not be flipped to positive, thus its negative impact on total sum is minimized by considering the smallest absolute value.

The method finally returns sumTotal as the maximum possible sum of the matrix values under the given conditions. This approach ensures that the sum is optimized by strategically managing negative values and leveraging the smallest value when necessary.

class Solution:
    def maximizeMatrixSum(self, mat: List[List[int]]) -> int:
        sum_absolute_values = 0
        smallest_value = float("inf")
        count_negatives = 0

        for row in mat:
            for item in row:
                sum_absolute_values += abs(item)
                if item < 0:
                    count_negatives += 1
                smallest_value = min(smallest_value, abs(item))

        # Modify sum if there is an odd number of negatives
        if count_negatives % 2 == 1:
            sum_absolute_values -= 2 * smallest_value

        return sum_absolute_values

The provided Python solution focuses on obtaining the maximum sum of a matrix where elements can be flipped from negative to positive. Here's the strategy detailed in the code:

1. Initialize variables to handle the total of absolute values (sum_absolute_values), the smallest absolute value encountered (smallest_value), and a count of negative numbers (count_negatives).

2. Traverse each element in the given matrix. During the traversal:

   • Add the absolute value of each element to sum_absolute_values.
   • Update count_negatives if the element is negative.
   • Determine smallest_value by comparing it to the absolute value of the current item.

3. Upon completion of the matrix traversal, check if count_negatives is odd. If it is:

   • Adjust sum_absolute_values by subtracting twice the smallest_value. This operation accounts for the possibility that one negative number could not be flipped (given an odd number of total negatives), and thus, the smallest absolute value should be subtracted from the total to maximize the sum.

4. Return sum_absolute_values as the result, which represents the maximum sum possible under the defined conditions.

This approach effectively uses mathematical analysis to optimize the matrix's sum by strategically managing negative values and maximizing the effective addition of all absolute values.