1175MediumArrayGreedyMatrix

Turning Signs in Pairs on the Grid

Tracked in this browser only

Trains the technique from

LeetCode 1975Maximum Matrix Sum

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A square grid grid holds whole numbers, some of them below zero. One move picks two cells sharing a side and turns the sign of both of them.

Return the largest total the grid's cells can add up to after any number of moves.

Examples

Example 1

Input: grid = [[-1, 1], [1, 1]]
Output: 2

One cell is below zero, and every move turns two signs, so the count of negatives can only change by two. One cell must stay negative, and the cheapest to leave is the smallest in size.

Example 2

Input: grid = [[0, -1], [1, 1]]
Output: 3

One cell is below zero again, but a cell holding nothing costs nothing to leave negative, so every other cell can be made positive.

Example 3

Input: grid = [[2, -3], [5, -7]]
Output: 17

Two cells are below zero, so both can be turned positive and the whole grid adds up in size.

Constraints

2 <= grid.length <= 250
grid[i].length == grid.length
-10^5 <= grid[i][j] <= 10^5

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def max_matrix_sum(grid: list[list[int]]) -> int:

Java

public long maxMatrixSum(int[][] grid)

The intended approach

Time O(n^2) over the cells.Space O(1).

A move turns two signs at once, so it never changes whether the number of cells below zero is odd or even. That parity is the only thing standing in the way, because any two cells of the grid can be joined by a chain of side-sharing steps, and turning the signs along such a chain flips exactly its two ends while every cell in between is flipped twice and left alone. So any two cells below zero can be cleared together. With an even count of negatives, every cell can therefore be made positive and the answer is the sizes all added up. With an odd count, one cell must stay below zero however the moves go, and the cheapest one to leave is the smallest in size, which costs twice that size against the sizes added up. A cell holding nothing counts as costing nothing to leave negative, which the same rule handles without a special case. The total runs past a thirty-two-bit integer at the allowed sizes.

Pythonentry point: max_matrix_sum

def max_matrix_sum(grid):
    total = 0
    negatives = 0
    smallest = None
    for row in grid:
        for value in row:
            size = value if value >= 0 else -value
            total += size
            if value < 0:
                negatives += 1
            if smallest is None or size < smallest:
                smallest = size
    if negatives % 2 == 1:
        total -= 2 * smallest
    return total

The obvious version

Walk outwards over every pattern of signs the moves can reach, starting from the grid as given and turning each pair of side-sharing cells in turn, keeping the largest total any reachable pattern gives. Costs two to the power of the cell count.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_max_matrix_sum(grid):
    from collections import deque
    n = len(grid)
    m = len(grid[0])
    flat = []
    for row in grid:
        for value in row:
            flat.append(value)
    start = tuple(flat)
    seen = set([start])
    queue = deque([start])
    best = None
    while queue:
        state = queue.popleft()
        total = 0
        for value in state:
            total += value
        if best is None or total > best:
            best = total
        for r in range(n):
            for c in range(m):
                for dr, dc in ((0, 1), (1, 0)):
                    nr = r + dr
                    nc = c + dc
                    if nr >= n or nc >= m:
                        continue
                    body = list(state)
                    a = r * m + c
                    b = nr * m + nc
                    body[a] = -body[a]
                    body[b] = -body[b]
                    nxt = tuple(body)
                    if nxt not in seen:
                        seen.add(nxt)
                        queue.append(nxt)
    return best

How this problem was checked

Reference solution executed against 15 test cases — 15 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

Create a free account to unlock the solution

See the step-by-step animation, the intuition, and clean code in every language — free, no credit card.

Create a free account Already have an account? Sign in