0783MediumArrayMatrixPrefix Sum

Neighbourhood Totals On A Sensor 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 rectangular sensor panel is given as a grid panel, where panel[i][j] is the reading of the sensor in row i, column j. Every row of the panel has the same length.

For each sensor the technician wants the combined reading of its neighbourhood. Return a grid out of the same shape as panel, where out[i][j] is the sum of panel[r][c] over every pair (r, c) that lies inside the panel and satisfies both

i - reach <= r <= i + reach, and
j - reach <= c <= j + reach.

The sensor at (i, j) is part of its own neighbourhood. Positions that fall off the edge of the panel do not exist and add nothing, so a neighbourhood near a corner simply holds fewer sensors. reach may be larger than the panel itself.

Examples

Example 1

Input: panel = [[2,5,1],[4,3,9],[7,6,8]], reach = 1
Output: [[14, 24, 18], [27, 45, 32], [20, 37, 26]]

For the corner cell (0, 0) the neighbourhood holds rows 0 to 1 and columns 0 to 1, so its total is 2 + 5 + 4 + 3 = 14. For the middle cell (1, 1) every sensor is within reach, giving 2 + 5 + 1 + 4 + 3 + 9 + 7 + 6 + 8 = 45.

Example 2

Input: panel = [[3,1,4,1],[5,9,2,6]], reach = 1
Output: [[18, 24, 23, 13], [18, 24, 23, 13]]

Cell (0, 2) covers both rows and columns 1 to 3, so its total is 1 + 4 + 1 + 9 + 2 + 6 = 23.

Example 3

Input: panel = [[6],[1],[9],[4],[2]], reach = 1
Output: [[7], [16], [14], [15], [6]]

The panel is a single column, so each neighbourhood is just the cell together with the ones directly above and below it where they exist: cell (0, 0) gives 6 + 1 = 7 and cell (2, 0) gives 1 + 9 + 4 = 14.

Constraints

1 <= panel.length <= 100
1 <= panel[i].length <= 100
1 <= reach <= 100
1 <= panel[i][j] <= 100
All rows of panel have equal length.

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 window_totals(panel: list[list[int]], reach: int) -> list[list[int]]:

Java

public int[][] windowTotals(int[][] panel, int reach)

The intended approach

Time O(m * n) for a grid of m rows and n columns: the table is filled once and every output cell costs a constant number of lookups.Space O(m * n) for the running-total table plus the returned grid.

Precompute a table whose entry `pre[r + 1][c + 1]` is the sum of the whole rectangle from the top-left corner of the panel down to `(r, c)`; it is filled row by row with a running row sum added to the row above. Any axis-aligned rectangle then costs four lookups, one per corner, with the two overlapping strips subtracted and the doubly-subtracted corner added back. For each output cell, clamp the block's row and column edges to the panel first, then read the rectangle out.

Pythonentry point: window_totals

def window_totals(panel, reach):
    # Build a running-total table where pre[r + 1][c + 1] holds the sum of every cell
    # above and to the left of (r, c) inclusive, then each answer is one rectangle
    # read out of it with four lookups after the block edges are clamped to the panel.
    # Time O(m * n), space O(m * n) for the table and the returned grid.
    m = len(panel)
    n = len(panel[0])
    pre = [[0] * (n + 1) for _ in range(m + 1)]
    for r in range(m):
        row = panel[r]
        above = pre[r]
        cur = pre[r + 1]
        run = 0
        for c in range(n):
            run += row[c]
            cur[c + 1] = above[c + 1] + run

    out = [[0] * n for _ in range(m)]
    for i in range(m):
        r1 = i - reach
        if r1 < 0:
            r1 = 0
        r2 = i + reach + 1
        if r2 > m:
            r2 = m
        top = pre[r1]
        bottom = pre[r2]
        for j in range(n):
            c1 = j - reach
            if c1 < 0:
                c1 = 0
            c2 = j + reach + 1
            if c2 > n:
                c2 = n
            out[i][j] = bottom[c2] - top[c2] - bottom[c1] + top[c1]
    return out

The obvious version

For each output cell, scan every sensor in the panel and add the ones whose row and column both sit within `reach` of that cell.

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_window_totals(panel, reach):
    m = len(panel)
    n = len(panel[0])
    out = []
    for i in range(m):
        row = []
        for j in range(n):
            total = 0
            for r in range(m):
                if abs(r - i) > reach:
                    continue
                for c in range(n):
                    if abs(c - j) <= reach:
                        total += panel[r][c]
            row.append(total)
        out.append(row)
    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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