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.
1def maxMatrixSum(matrix):
2 n = len(matrix)
3 min_abs = abs(matrix[0][0])
4 sum_ = 0
5 neg_count = 0
6 for i in range(n):
7 for j in range(n):
8 if matrix[i][j] < 0:
9 neg_count += 1
10 sum_ += abs(matrix[i][j])
11 min_abs = min(min_abs, abs(matrix[i][j]))
12 if neg_count % 2 != 0:
13 sum_ -= 2 * min_abs
14 return sum_The Python code follows similar logic, scanning the matrix to determine the total sum and number of negatives, then adjusting by twice the minimum absolute value if negatives are odd to maximize the 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.
1function maxMatrixSumOptimized(matrix) {
2 let n = matrix.length;
3 let minAbsVal = Math.abs(matrix[0][0]);
4 let totalSum = 0;
5 let negativeCount = 0;
6 for (let i = 0; i < n; i++) {
7 for (let j = 0; j < n; j++) {
8 let valueAbs = Math.abs(matrix[i][j]);
9 totalSum += valueAbs;
10 if (matrix[i][j] < 0) negativeCount++;
11 minAbsVal = Math.min(minAbsVal, valueAbs);
12 }
13 }
14 if (negativeCount % 2 !== 0) {
15 totalSum -= 2 * minAbsVal;
16 }
17 return totalSum;
18}For JavaScript, dynamic count and absolute summation of items alongside directional reasoning enable value computing. Allow adjustment via small absolute if parity falls into an unbalanced tally after entirely parsing.