To maximize the sum of the matrix, we should aim to minimize the negative influence. Notice that flipping any pair of adjacent elements will keep the number of negative elements consistent in terms of parity (even or odd). Therefore, the key is to ensure the sum of the negative values is minimized while keeping the matrix's parity even, allowing us to theoretically flip elements to achieve a positive parity. It implies that to maximize the sum, the total count of negative numbers should remain even by flipping the sign of the element with the smallest absolute value.
Time Complexity: O(n^2) because we need to iterate over each element in the matrix.
Space Complexity: O(1) as we use a constant amount of additional space.
1function maxMatrixSum(matrix) {
2 let n = matrix.length;
3 let minAbs = Math.abs(matrix[0][0]);
4 let sum = 0;
5 let negCount = 0;
6 for (let i = 0; i < n; i++) {
7 for (let j = 0; j < n; j++) {
8 if (matrix[i][j] < 0) negCount++;
9 sum += Math.abs(matrix[i][j]);
10 minAbs = Math.min(minAbs, Math.abs(matrix[i][j]));
11 }
12 }
13 if (negCount % 2 !== 0) {
14 sum -= 2 * minAbs;
15 }
16 return sum;
17}This JavaScript approach mirrors the language above where the iteration gathers the total sum and negative count, adjusting for an odd number of negatives utilizing twice the minimum absolute value upon completion to deliver a mathematically maximal sum.
This approach further explores the combination of summing absolute values and preserving parity. Since the act of flipping adjacent pairs maintains the overall parity of negatives, the challenge is to manage the balance of signs dynamically while ensuring operations help convert the majority of negative numbers or maintain equilibrium by pairing onto the minimal interference possible.
Time Complexity: O(n^2); the necessity exists in exploring each point.
Space Complexity: O(1); friendly constant load aside typical iteration encapsulation.
1def maxMatrixSumOptimized(matrix):
2 n = len(matrix)
3 min_abs = abs(matrix[0][0])
4 total_sum = 0
5 neg_count = 0
6 for i in range(n):
7 for j in range(n):
8 val_abs = abs(matrix[i][j])
9 total_sum += val_abs
10 if matrix[i][j] < 0:
11 neg_count += 1
12 min_abs = min(min_abs, val_abs)
13 if neg_count % 2 != 0:
14 total_sum -= 2 * min_abs
15 return total_sumPythons' path again involves a straightforward count alongside an abs sum tally during navigation, adjusting by the minimal absolute if parity from negative numbers ends unequally, ensuring optimal matrix positivity.