0641EasyArrayMatrix

Softening A Grey Level Scan

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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 scanner returns img, a grid of grey levels with m rows and n columns. img[r][c] is the grey level recorded at row r, column c.

Softening the scan replaces the level at each cell with the average level over a block: the cell itself together with every cell that touches it, sideways, vertically or diagonally, so at most nine cells. Positions that would sit outside the grid are left out of both the sum and the count. Round each average down to a whole number. Return the softened grid, which has the same shape as img. Every average is taken from the original levels, not from levels already softened.

Examples

Example 1

Input: img = [[1, 2, 3, 4]]
Output: [[1, 2, 3, 3]]

The grid has one row. Column 0 averages 1 and 2, which rounds down to 1. Column 1 averages 1, 2 and 3, giving 2. Column 2 averages 2, 3 and 4, giving 3. Column 3 averages 3 and 4, which rounds down to 3.

Example 2

Input: img = [[10], [20], [30]]
Output: [[15], [20], [25]]

The grid has one column. Row 0 averages 10 and 20, giving 15. Row 1 averages all three levels, giving 20. Row 2 averages 20 and 30, giving 25.

Example 3

Input: img = [[8, 3, 9], [1, 7, 2], [6, 4, 5], [2, 2, 2]]
Output: [[4, 5, 5], [4, 5, 5], [3, 3, 3], [3, 3, 3]]

The corner cell at row 0 column 0 averages the four levels 8, 3, 1 and 7, which is 4 after rounding down. The interior cell at row 1 column 1 averages all nine levels 8, 3, 9, 1, 7, 2, 6, 4 and 5, which is 5 after rounding down.

Constraints

m == img.length
n == img[i].length
1 <= m, n <= 200
0 <= img[i][j] <= 255

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 image_smoother(img: list[list[int]]) -> list[list[int]]:

Java

public int[][] imageSmoother(int[][] img)

The intended approach

Time O(m * n) — at most nine reads per cell.Space O(m * n) for the softened grid that is returned.

Allocate a second grid of the same shape and fill it cell by cell. For a cell, run over the three candidate rows and three candidate columns around it, skip any that fall outside the grid, and keep a running sum and a running count of the ones that do not. Integer division of the sum by the count rounds the average down. Results are written to the second grid so that a softened level is never read back as if it were an original one.

Pythonentry point: image_smoother

def image_smoother(img):
    # O(m * n) time with nine reads per cell, O(m * n) for the grid returned.
    # Every average is taken from the original grid, so results are written into
    # a fresh grid rather than back into img.
    m = len(img)
    n = len(img[0])
    out = [[0] * n for _ in range(m)]
    for r in range(m):
        for c in range(n):
            total = 0
            count = 0
            for rr in range(r - 1, r + 2):
                if rr < 0 or rr >= m:
                    continue
                row = img[rr]
                for cc in range(c - 1, c + 2):
                    if cc < 0 or cc >= n:
                        continue
                    total += row[cc]
                    count += 1
            out[r][c] = total // count
    return out

The obvious version

Build a two-dimensional running-sum table over the original grid, then for each cell clamp the surrounding block to the grid boundary, read that block's sum straight out of the table with four lookups, and divide by the number of cells the clamped block covers.

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_image_smoother(img):
    m = len(img)
    n = len(img[0])
    pre = [[0] * (n + 1) for _ in range(m + 1)]
    for r in range(m):
        for c in range(n):
            pre[r + 1][c + 1] = pre[r][c + 1] + pre[r + 1][c] - pre[r][c] + img[r][c]
    out = []
    for r in range(m):
        line = []
        for c in range(n):
            r0 = r - 1 if r - 1 > 0 else 0
            c0 = c - 1 if c - 1 > 0 else 0
            r1 = r + 1 if r + 1 < m - 1 else m - 1
            c1 = c + 1 if c + 1 < n - 1 else n - 1
            block = (pre[r1 + 1][c1 + 1] - pre[r0][c1 + 1]
                     - pre[r1 + 1][c0] + pre[r0][c0])
            cells = (r1 - r0 + 1) * (c1 - c0 + 1)
            line.append(block // cells)
        out.append(line)
    return out

How this problem was checked

Reference solution executed against 10 test cases — 10 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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