0920MediumArraySortingMatrix

Closest Distinct Pair in Each Window

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Trains the technique from

LeetCode 3567Minimum Absolute Difference in Sliding Submatrix

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 board of readings is given as grid.

For every k by k window of the board, report the smallest positive difference between two readings inside it, or 0 when every reading in that window is the same.

Return those figures as a board of their own, laid out so that the entry at row i and column j answers the window whose top-left corner sits at row i, column j.

Examples

Example 1

Input: grid = [[14, 3, 27], [9, 21, 6], [17, 8, 31]], k = 2
Output: [[5, 3], [1, 2]]

The top-left window holds 14, 3, 9 and 21; in order those are 3, 9, 14 and 21, and the closest neighbours are 9 and 14, five apart. The other three windows are read the same way.

Example 2

Input: grid = [[7, 7], [7, 7]], k = 2
Output: [[0]]

The single window holds one distinct reading, so there is no pair to measure.

Example 3

Input: grid = [[1, 2, 3], [4, 5, 6], [7, 8, 9]], k = 1
Output: [[0, 0, 0], [0, 0, 0], [0, 0, 0]]

Every window is a single cell, so none of them holds a pair.

Constraints

1 <= grid.length <= 30
1 <= grid[i].length <= 30
Every row of grid has the same length
-10^5 <= grid[i][j] <= 10^5
1 <= k <= min(grid.length, grid[i].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 min_abs_diff(grid: list[list[int]], k: int) -> list[list[int]]:

Java

public int[][] minAbsDiff(int[][] grid, int k)

The intended approach

Time O(m * n * k^2 log(k^2)) for an m by n board and windows of side k.Space O(k^2) for one window's readings.

Inside one window, the closest two distinct readings must sit next to each other once the distinct readings are put in order, because any other pair straddles at least one of those neighbouring gaps and is therefore no closer. So gather the window's readings, drop the repeats, sort what is left, and take the smallest step between neighbours. A window down to a single distinct reading has no pair at all and reports nothing. The board is small enough that every window can be gathered outright, so the whole answer is one sweep over the window positions with a sort inside each.

Pythonentry point: min_abs_diff

def min_abs_diff(grid, k):
    m = len(grid)
    n = len(grid[0])
    out = []
    for top in range(m - k + 1):
        row = []
        for left in range(n - k + 1):
            seen = set()
            for r in range(top, top + k):
                for c in range(left, left + k):
                    seen.add(grid[r][c])
            values = sorted(seen)
            if len(values) < 2:
                row.append(0)
                continue
            closest = values[1] - values[0]
            for i in range(2, len(values)):
                step = values[i] - values[i - 1]
                if step < closest:
                    closest = step
            row.append(closest)
        out.append(row)
    return out

The obvious version

For every window, compare every pair of its readings directly and keep the smallest difference that is not zero. Quartic in the window side and it never sorts anything.

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_min_abs_diff(grid, k):
    m = len(grid)
    n = len(grid[0])
    out = []
    for top in range(m - k + 1):
        row = []
        for left in range(n - k + 1):
            cells = []
            for r in range(top, top + k):
                for c in range(left, left + k):
                    cells.append(grid[r][c])
            closest = 0
            for i in range(len(cells)):
                for j in range(i + 1, len(cells)):
                    step = cells[i] - cells[j]
                    if step < 0:
                        step = -step
                    if step > 0 and (closest == 0 or step < closest):
                        closest = step
            row.append(closest)
        out.append(row)
    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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