0520MediumArrayDynamic ProgrammingMatrix

Strongest Signal Across the Amplifier Grid

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Trains the technique from

LeetCode 1594Maximum Non Negative Product in a Matrix

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 test rig is a rectangular board of amplifier tiles, given as grid, where grid[r][c] is the gain of the tile in row r and column c. A gain may be negative, which means that tile inverts the signal.

A probe signal enters at the top-left tile with value 1 and must leave at the bottom-right tile. From a tile the probe may only step one tile to the right or one tile down. Every tile the probe occupies, including the entry tile and the exit tile, multiplies the running value by that tile's gain.

Among the routes whose final value is not negative, report the largest final value. Because that value can be enormous, report it modulo 1000000007. If every route ends on a negative value, return -1 instead.

Examples

Example 1

Input: grid = [[-2, 3], [4, -1]]
Output: 8

Stepping down and then right puts the probe on -2, 4 and -1, so the value leaving the board is 8. That is not negative, and 8 reduced modulo 1000000007 is 8.

Example 2

Input: grid = [[2, -1, -1, 3, 1, 2]]
Output: 12

The board is a single row, so the probe walks every tile: 2, -1, -1, 3, 1 and 2 multiply to 12.

Example 3

Input: grid = [[-1, 2], [3, 4]]
Output: -1

The two available routes leave the board holding -8 and -12. Neither is allowed, so the answer is -1.

Constraints

rows == grid.length
cols == grid[r].length
1 <= rows, cols <= 15
-4 <= grid[r][c] <= 4
The reported value is the largest non-negative final value reduced modulo 1000000007, or -1.

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_product_path(grid: list[list[int]]) -> int:

Java

public int maxProductPath(int[][] grid)

The intended approach

Time O(rows * cols), a constant amount of work per tile.Space O(rows * cols) for the two boards of running values.

Sweep the board once in row-major order, keeping two running values per tile: the largest value that can arrive there and the smallest. A negative gain turns the smallest arriving value into the largest leaving one, so both have to travel together; each tile combines the two values held by the tile above and the two held by the tile to the left with its own gain and keeps the extremes. The reduction modulo 1000000007 is applied only to the final answer, so the comparisons all run on true values.

Pythonentry point: max_product_path

def max_product_path(grid):
    # O(rows*cols) time: each tile keeps the largest and the smallest running
    # value that can reach it, because one more negative factor swaps the two.
    MOD = 1000000007
    rows = len(grid)
    cols = len(grid[0])
    hi = [[0] * cols for _ in range(rows)]
    lo = [[0] * cols for _ in range(rows)]
    hi[0][0] = grid[0][0]
    lo[0][0] = grid[0][0]
    for c in range(1, cols):
        hi[0][c] = hi[0][c - 1] * grid[0][c]
        lo[0][c] = lo[0][c - 1] * grid[0][c]
    for r in range(1, rows):
        hi[r][0] = hi[r - 1][0] * grid[r][0]
        lo[r][0] = lo[r - 1][0] * grid[r][0]
    for r in range(1, rows):
        for c in range(1, cols):
            f = grid[r][c]
            a = hi[r - 1][c] * f
            b = lo[r - 1][c] * f
            d = hi[r][c - 1] * f
            e = lo[r][c - 1] * f
            hi[r][c] = max(a, b, d, e)
            lo[r][c] = min(a, b, d, e)
    best = hi[rows - 1][cols - 1]
    if best < 0:
        return -1
    return best % MOD

The obvious version

Carry the full set of values reachable at each tile rather than just the extremes: a tile's set is every value in the set above and every value in the set to the left, each multiplied by the tile's gain. Then scan the exit tile's set for the largest entry that is not negative.

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_product_path(grid):
    MOD = 1000000007
    rows = len(grid)
    cols = len(grid[0])
    reach = [[None] * cols for _ in range(rows)]
    reach[0][0] = {grid[0][0]}
    for r in range(rows):
        for c in range(cols):
            if r == 0 and c == 0:
                continue
            pool = set()
            if r > 0:
                for v in reach[r - 1][c]:
                    pool.add(v * grid[r][c])
            if c > 0:
                for v in reach[r][c - 1]:
                    pool.add(v * grid[r][c])
            reach[r][c] = pool
    winner = None
    for v in reach[rows - 1][cols - 1]:
        if v >= 0 and (winner is None or v > winner):
            winner = v
    if winner is None:
        return -1
    return winner % MOD

How this problem was checked

Reference solution executed against 11 test cases — 11 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

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