0048MediumArrayMathMatrix

Wafer Quarter Turn

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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.

An inspection rig stores one calibration offset per die of a wafer in grid, a square board of n rows by n columns. Offsets are signed: a die can read below the reference as easily as above it.

The wafer has just been remounted a quarter turn clockwise, so the stored board no longer lines up with the hardware. Turn the board itself a quarter turn clockwise: the row that was on top must end up as the rightmost column, read downwards, and so on around the board.

The rig has no spare memory for a second board of this size, so you must shuffle the values inside grid where they already sit rather than filling in a freshly allocated board and handing that back. Only a fixed number of scratch values is allowed, no matter how large n is. After rearranging grid, return grid.

Boards that are not square are outside the scope of this task, so you may assume every row is exactly as long as the number of rows.

Examples

Example 1

Input: grid = [[3, -7], [12, 0]]
Output: [[12, 3], [0, -7]]

The left column 3, 12 becomes the top row read upwards, so 12 lands first and 3 follows.

Example 2

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

Each column, read from the bottom up, becomes a row of the turned board. The centre die never moves.

Example 3

Input: grid = [[4]]
Output: [[4]]

A single die is its own row and column, so a quarter turn leaves the board as it was.

Constraints

n == grid.length == grid[i].length
1 <= n <= 20
-1000 <= grid[i][j] <= 1000
Only a fixed amount of extra space may be used; no second board of size n by n

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 quarter_turn(grid: list[list[int]]) -> list[list[int]]:

Java

public int[][] quarterTurn(int[][] grid)

The intended approach

Time O(n^2)Space O(1)

Peel the board into concentric rings and rotate each ring on the spot. For every position along the top side of a ring, hold that value aside and then move three values in one chain: left side into top, bottom into left, right into bottom, and finally the held value into the right. Two nested loops cover each ring exactly once, and the only scratch used is a single held value.

Pythonentry point: quarter_turn

def quarter_turn(grid):
    n = len(grid)
    for layer in range(n // 2):
        last = n - 1 - layer
        for i in range(layer, last):
            offset = i - layer
            held = grid[layer][i]
            grid[layer][i] = grid[last - offset][layer]
            grid[last - offset][layer] = grid[last][last - offset]
            grid[last][last - offset] = grid[i][last]
            grid[i][last] = held
    return grid

The obvious version

Ignore the space rule and fill a brand new board from the index identity: whatever sits at row i column j belongs at row j column n-1-i after a clockwise quarter turn. Direct index mapping, no swap chains.

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_quarter_turn(grid):
    n = len(grid)
    turned = [[0] * n for _ in range(n)]
    for i in range(n):
        for j in range(n):
            turned[j][n - 1 - i] = grid[i][j]
    return turned

How this problem was checked

Reference solution executed against 8 test cases — 8 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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