0199MediumArrayMatrixSimulation

Slanted Sweep Readout

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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 rig stores its readings in a grid plate with m rows and n columns, where plate[i][j] is a signed offset in micrometres and may be negative.

The rig's arm does not read row by row. It groups the cells into sweeps: two cells belong to the same sweep when their row index and column index add up to the same total. The arm works through the sweeps in increasing order of that total, starting with the sweep whose total is 0.

Inside a sweep the arm alternates which way it travels. When the sweep's total is even, the arm reports its cells starting from the largest row index and working down to the smallest. When the total is odd, it reports them from the smallest row index up to the largest.

Return the offsets in the order the arm reports them.

Examples

Example 1

Input: plate = [[-4, 7, 2], [5, -1, 9]]
Output: [-4, 7, 5, -1, 2, 9]

Sweep 0 holds only -4. Sweep 1 has an odd total so it runs downward through 7 then 5. Sweep 2 is even and runs upward through -1 then 2, and sweep 3 finishes with 9.

Example 2

Input: plate = [[8], [-3], [6]]
Output: [8, -3, 6]

With a single column every sweep holds exactly one cell, so the direction never matters and the column is read top to bottom.

Example 3

Input: plate = [[9, -9]]
Output: [9, -9]

A single row also puts one cell in each sweep, so the readings come out left to right.

Constraints

m == plate.length
n == plate[i].length
1 <= m, n <= 10^4
1 <= m * n <= 10^4
-10^5 <= plate[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 slanted_sweep_readout(plate: list[list[int]]) -> list[int]:

Java

public int[] slantedSweepReadout(int[][] plate)

The intended approach

Time O(m * n)Space O(1) extra beyond the returned list

Index the sweeps by the shared total of row and column, which runs from 0 to m + n - 2. For a given total the row index is bounded below by what keeps the column inside the grid and above by the last row, so the whole sweep can be walked directly without collecting it first. Walk that row range downward on even totals and upward on odd ones, deriving the column as the total minus the row.

Pythonentry point: slanted_sweep_readout

def slanted_sweep_readout(plate):
    # Time: O(m * n) -- every cell is visited exactly once across all sweeps.
    # Space: O(1) beyond the output, since no sweep is buffered before emitting.
    rows = len(plate)
    cols = len(plate[0])
    out = []
    for total in range(rows + cols - 1):
        low = max(0, total - cols + 1)
        high = min(total, rows - 1)
        order = range(high, low - 1, -1) if total % 2 == 0 else range(low, high + 1)
        for r in order:
            out.append(plate[r][total - r])
    return out

The obvious version

Simulate the arm one cell at a time. Hold a current position and a travel direction, emit the cell under the arm, then step diagonally, and when a step would leave the grid, slide to the next sweep and flip the direction. No sweep index arithmetic.

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_slanted_sweep_readout(plate):
    rows = len(plate)
    cols = len(plate[0])
    r = 0
    c = 0
    going_up = True
    out = []
    for _ in range(rows * cols):
        out.append(plate[r][c])
        if going_up:
            if c == cols - 1:
                r += 1
                going_up = False
            elif r == 0:
                c += 1
                going_up = False
            else:
                r -= 1
                c += 1
        else:
            if r == rows - 1:
                c += 1
                going_up = True
            elif c == 0:
                r += 1
                going_up = True
            else:
                r += 1
                c -= 1
    return out

How this problem was checked

Reference solution executed against 12 test cases — 12 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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