1975. Maximum Matrix Sum - Detailed Explanation

Explanation: This optimal solution scans the matrix once (O(n^2) time). It calculates the sum of absolute values (treating every number as positive), counts how many negatives are present, and finds the smallest absolute value. Then it adjusts the sum if needed. If there’s an odd number of negatives (and no zero to offset that), it subtracts 2 * min_abs_val from the total absolute sum – this is equivalent to forcing one smallest-magnitude number to be negative. This approach is greedy but it can be proven to always yield the optimal result for this problem.

Complexity Analysis

• Brute Force Approach: Time complexity is exponential, as we might need to examine on the order of 2^K states (where K is the number of possible flip operations or effectively the number of edges in the adjacency graph). Even for small matrices this grows very fast. For example, a 3x3 matrix has 12 possible adjacent pairs, yielding up to 2^12 = 4096 combinations to consider (many leading to duplicate states). The space complexity is also exponential in the worst case due to storing a large number of matrix states in the search. This approach is more of a conceptual check and is not feasible for large n.

• Optimal Greedy Approach: Time complexity is O(n^2), since we just need to iterate through the matrix once to compute sums and counts. For n up to 250, this is at most 62,500 elements, which is very efficient. Space complexity is O(1) extra space, since we use only a few counters/variables and do in-place calculations. (We are not using additional structures proportional to n, aside from output storage or the input itself).

In practice, the optimal solution is extremely fast even for the largest allowed matrix, whereas the brute force would run out of time or memory for anything beyond the tiniest matrices.

Common Mistakes

• Ignoring the Flip Constraint: A common mistake is to assume you can flip each element independently. Simply turning all negatives to positive (i.e., summing absolute values) without considering parity might give a sum that isn’t actually achievable under the rules. Always remember that flips occur in pairs.

• Forgetting the Parity Rule: Not recognizing that an odd number of negatives forces one negative to remain (when no zero is present) can lead to an incorrect solution. Some might incorrectly return total_abs_sum in cases where there’s an odd count of negatives but no zero, which would be wrong.

• Not Accounting for Zero: If there’s a zero in the matrix, it’s a game-changer. Forgetting to handle the zero case could cause you to unnecessarily subtract the smallest value even when it’s not needed. (In the optimal approach code above, this is handled implicitly because if a zero exists, min_abs_val would be 0, making the subtraction effectively zero. Alternatively, one could explicitly check for zero.)

• Using Wrong Data Types: The sum of absolute values can be large (up to n*n*100,000). For n=250, the maximum sum could be 250250100000 = 6.25 × 10^9, which exceeds the range of a 32-bit integer. In languages like Java or C++, using an int for the sum might overflow. Make sure to use a larger type (long in Java) for summing to avoid overflow errors.

• Double Counting or Missing Elements: When implementing, be careful to include every element’s absolute value in the sum and check every element for negativity and for the smallest absolute. Skipping an element or counting something twice could bug the result.

• Flipping the Wrong Element in Odd Case: If you do realize one element must remain negative, ensure it’s the one with the smallest absolute value globally, not just the smallest negative. For instance, if the smallest absolute value in the matrix is a positive number, you might want to sacrifice that one (by flipping it to negative via another operation) instead of keeping a larger-magnitude negative. The strategy should be global, not local per row/column.

Edge Cases to Consider

• All Positive Matrix: If every element is already positive (or zero), then no flips are needed at all. The answer is just the sum of all elements. (Our formula would yield neg_count = 0 which is even, so total_abs_sum is returned, correctly handling this case).

• All Negative Matrix: If all numbers are negative:

  • If the count of negatives is even, you can flip them in pairs to make all of them positive. The result will be the sum of absolute values of all numbers.
  • If the count of negatives is odd (and there’s no zero), one number will remain negative. For example, in a 3x3 all-negative matrix with an odd total count, the answer will be sum(|nums|) - 2 * min(|nums|) because one smallest-magnitude number stays negative.

• Matrix with Zeroes: If any zero is present, you can always achieve the absolute sum of all numbers, even if there’s an odd number of negatives. The zero can be paired with any one negative to flip it to positive without introducing a new negative. For example, in a matrix [[ -5, 2], [0, 3]] (odd negatives count, but a zero present), the maximum sum would be | -5| + |2| + |0| + |3| = 10 (we can flip -5 with 0 to get 5 and 0).

• Mix of Positives and Negatives: Ensure the logic covers scenarios like one negative among many positives, or one positive among many negatives. For instance, [[ -5, 2], [3, 4]] (one negative) should result in flipping that -5 with the smallest positive (2) to yield a negative 2 and positive 5, leaving matrix [[-2, -?], ...] etc. The optimal sum ends up with one negative (the 2) and is computed correctly by the formula.

• Duplicate Minimum Values: If the smallest absolute value appears multiple times, any one of them being negative in the end yields the same result. The algorithm just needs one representative smallest value. (E.g., if two cells have value 1 and others are larger, and you have odd negatives, only one of those 1’s will end up negative – sum impact is the same no matter which one).

• Large Magnitudes: Check that using the smallest value logic indeed picks the right candidate. If the smallest absolute value is, say, 0 (meaning a zero is present), the formula correctly doesn’t subtract anything. If it’s a very small positive number, the algorithm will choose to make that one negative if needed.

Alternative Variations

This problem can be adapted or viewed in different ways:

• Single Element Flips: If the operation allowed flipping a single element’s sign (instead of two adjacent), the solution would be trivial – just flip every negative to positive. The interesting part of this problem is the adjacency (pair-flip) constraint which introduces the parity consideration.

• Flipping Whole Rows/Columns: Consider a variation where instead of adjacent pairs, you could flip an entire row or entire column. This becomes a different puzzle (related to matrix parity and often solvable by greedily flipping rows with negative values). In fact, LeetCode’s “Score After Flipping Matrix” (problem 861) deals with flipping entire rows or columns of a binary matrix to maximize a binary number interpretation. The underlying idea is also to maximize the value by making most entries “positive” (1 in that case).

• Limited Number of Flips: Another variant could be if you were only allowed a fixed number of flip operations (say at most K flips). Then you’d have to be strategic about which pairs to flip first (likely flip the largest-magnitude negatives first). That would turn it into an optimization problem with a budget on operations.

• Different Adjacency Rules: Instead of flipping adjacent orthogonally, one could imagine a twist where any two elements (not necessarily adjacent) could be flipped together. That would actually simplify the problem because any negative could pair with any other negative or with the smallest positive, making it always possible to maximize sum similarly to our solution. The adjacency requirement in this problem is what ties the parity of negatives.

• Minimizing Sum: As a spin-off, one could ask to minimize the matrix sum with similar operations. The strategy would then be the opposite: we’d want as many negatives as possible. By a similar parity argument, you’d try to make everything negative (take negatives of all abs values) unless prevented by parity (odd number of positives means one stays positive).

• Higher Dimensions or Non-Square Grids: The concept could extend to 3D grids or non-square matrices (though the problem as stated requires a square matrix). The parity argument would remain similar: flipping an even number of negatives keeps parity, etc., just extended to that structure.

Each of these variations would require adjusting the approach, but understanding the parity and absolute value strategy from this problem provides a strong foundation for tackling them.

Related Problems for Further Practice

• LeetCode 1005 – Maximize Sum of Array After K Negations: A one-dimensional analog of this problem. You are given an array and can flip the sign of at most K elements. The goal is to maximize the array sum. This can be solved greedily by repeatedly flipping the smallest negative value or, if no negatives left, possibly flipping the smallest positive (or zero) if you still have flips left.

• LeetCode 861 – Score After Flipping Matrix: A greedy matrix problem where you can flip any row or any column (invert all bits in that row/column) of a binary matrix. The aim is to maximize the “score” of the matrix where each row is interpreted as a binary number. It also involves making decisions to maximize 1s (positives in a sense) in certain positions.

• LeetCode 832 – Flipping an Image: In this problem, you flip each row of a binary matrix horizontally and then invert the bits (0->1, 1->0). It’s a simpler transformation problem, but it gets you comfortable with the idea of flipping matrix entries and thinking about how operations affect the whole matrix.

• LeetCode 1252 – Cells with Odd Values in a Matrix: This involves adding 1 to all cells in a row or column (an operation that affects parity). While not about maximizing a sum, it explores the effect of toggling operations on a matrix and counting outcomes (in that case, how many cells end up odd). It’s another perspective on applying operations across rows/cols and their cumulative effect.