1117EasyArrayMatrixSimulation

Laying the Tiles Into a Rack

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Trains the technique from

LeetCode 2022Convert 1D Array Into 2D Array

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 run of tiles is to be laid into a rack of rows shelves, each shelf holding cols slots. Fill the rack shelf by shelf from the top, and within each shelf from left to right, taking the tiles in the order they are given.

Return the filled rack. When the run of tiles does not fill the rack exactly, return an empty list.

Examples

Example 1

Input: tiles = [1, 2, 3, 4, 5, 6], rows = 2, cols = 3
Output: [[1, 2, 3], [4, 5, 6]]

Six tiles fill two shelves of three slots, taken in order along the top shelf and then the bottom one.

Example 2

Input: tiles = [1, 2, 3, 4, 5], rows = 2, cols = 2
Output: []

Five tiles cannot fill a rack of four slots, whichever way they are laid, so nothing is returned.

Example 3

Input: tiles = [5, 4, 3, 2, 1], rows = 5, cols = 1
Output: [[5], [4], [3], [2], [1]]

Five shelves of a single slot take the tiles one at a time in the order given.

Constraints

1 <= tiles.length <= 5 * 10^4
1 <= tiles[i] <= 10^5
1 <= rows <= 4 * 10^4
1 <= cols <= 4 * 10^4

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 construct2_d_array(tiles: list[int], rows: int, cols: int) -> list[list[int]]:

Java

public int[][] construct2DArray(int[] tiles, int rows, int cols)

The intended approach

Time O(rows * cols) to copy every tile once.Space O(rows * cols) for the rack.

The rack holds `rows` times `cols` slots, so the only run of tiles that fits is one of exactly that length. Any other length fails outright, whether it is short or long, and that check comes first because multiplying the two sides can be done in one step while laying out tiles cannot be undone. Once the length matches, the filling order means shelf `r` takes the tiles from position `r * cols` up to just before `(r + 1) * cols`. Cutting the run into consecutive blocks of `cols` tiles walks each shelf in turn without any index arithmetic per slot.

Pythonentry point: construct2_d_array

def construct2_d_array(tiles, rows, cols):
    if len(tiles) != rows * cols:
        return []
    rack = []
    at = 0
    for _ in range(rows):
        rack.append(tiles[at:at + cols])
        at += cols
    return rack

The obvious version

Walk the rack slot by slot and work out, for each shelf and slot, which position of the run belongs there by multiplying the shelf number by the slot count and adding the slot. Costs one lookup per slot rather than one slice per shelf.

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_construct2_d_array(tiles, rows, cols):
    if len(tiles) != rows * cols:
        return []
    rack = []
    for r in range(rows):
        shelf = []
        for c in range(cols):
            shelf.append(tiles[r * cols + c])
        rack.append(shelf)
    return rack

How this problem was checked

Reference solution executed against 15 test cases — 15 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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