0832MediumArrayMatrixPrefix Sum

Cross Product Panel

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Trains the technique from

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 calibration panel is a grid of n rows by m columns of readings, given as grid.

Build a second grid panel of the same shape, where panel[i][j] is the product of every reading of grid except the one at row i, column j, taken modulo 12345.

Return panel.

Examples

Example 1

Input: grid = [[4, 7], [2, 5]]
Output: [[70, 40], [140, 56]]

For the cell holding 4 the other readings are 7, 2 and 5, whose product is 70. For the cell holding 7 they are 4, 2 and 5, giving 40. For 2 they are 4, 7 and 5, giving 140, and for 5 they are 4, 7 and 2, giving 56. All four are below 12345, so the remainder leaves them unchanged.

Example 2

Input: grid = [[6], [7]]
Output: [[7], [6]]

The panel has one column. The cell holding 6 has only the 7 left, and the cell holding 7 has only the 6.

Example 3

Input: grid = [[12345, 4]]
Output: [[4, 0]]

For the first cell the remaining product is 4. For the second it is 12345, and 12345 leaves a remainder of 0 on division by 12345.

Constraints

1 <= grid.length <= 10^5
1 <= grid[0].length <= 10^5
Every row of grid has the same length, and grid holds at least 2 readings in total
1 <= grid[i][j] <= 10^9

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

Java

public int[][] crossProductGrid(int[][] grid)

The intended approach

Time O(n * m): two sweeps of the cells.Space O(n * m) for the forward products, plus the returned grid.

Read the grid as one long run of cells, row after row. Sweep forward keeping the running product of everything seen so far, reduced each step, which gives the product of everything before each cell. Sweep back keeping the running product of everything after each cell, and multiply the two. Division is never used, which matters because the answer is a remainder and dividing a remainder does not recover the quotient.

Pythonentry point: cross_product_grid

def cross_product_grid(grid):
    MOD = 12345
    rows = len(grid)
    cols = len(grid[0])
    total = rows * cols
    # running product of everything before each cell, then everything after it
    before = [1] * (total + 1)
    flat = []
    for row in grid:
        flat.extend(row)
    for i in range(total):
        before[i + 1] = (before[i] * flat[i]) % MOD
    after = 1
    out = [[0] * cols for _ in range(rows)]
    for i in range(total - 1, -1, -1):
        out[i // cols][i % cols] = (before[i] * after) % MOD
        after = (after * flat[i]) % MOD
    return out

The obvious version

For each cell, walk every other cell of the grid and multiply, reducing as it goes. Quadratic in the number of cells, but it applies the definition literally and never reuses a partial product.

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_cross_product_grid(grid):
    MOD = 12345
    rows = len(grid)
    cols = len(grid[0])
    out = []
    for r in range(rows):
        line = []
        for c in range(cols):
            product = 1
            for rr in range(rows):
                for cc in range(cols):
                    if rr == r and cc == c:
                        continue
                    product = (product * grid[rr][cc]) % MOD
            line.append(product)
        out.append(line)
    return out

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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