0589EasyArrayMatrixSimulation

Flip a Reading Grid Onto Its Side

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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 logging rig writes its readings into matrix, a rectangle with one row per sensor and one column per sample instant.

The analysis stage wants the same readings laid out the other way round: one row per sample instant and one column per sensor. Mirror the rectangle across the line running from its top-left corner down and to the right, so the reading stored at row r, column c ends up at row c, column r.

Return the mirrored rectangle.

Examples

Example 1

Input: matrix = [[9, -4, 6], [2, 8, -1]]
Output: [[9, 2], [-4, 8], [6, -1]]

The rig had two sensors and three sample instants, so the answer has three rows of two readings. The reading 6 sat at row 0, column 2 and now sits at row 2, column 0.

Example 2

Input: matrix = [[4], [5], [6]]
Output: [[4, 5, 6]]

Three sensors sampled once each. Mirroring turns the single column into a single row.

Example 3

Input: matrix = [[2, 6], [8, 3]]
Output: [[2, 8], [6, 3]]

The two readings on the mirror line, 2 and 3, stay where they are, and 6 and 8 trade places.

Constraints

m == matrix.length
n == matrix[i].length
1 <= m, n <= 1000
1 <= m * n <= 10^5
-10^9 <= matrix[i][j] <= 10^9

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 transpose(matrix: list[list[int]]) -> list[list[int]]:

Java

public int[][] transpose(int[][] matrix)

The intended approach

Time O(m * n) — one visit per reading.Space O(m * n) for the rectangle handed back, O(1) beyond it.

The mirrored rectangle has `n` rows and `m` columns, so allocate that shape first and fill it in place of growing lists. Then walk the input row by row, which keeps each source row in cache, and write every reading to the slot whose row and column indices are the swap of its own.

Pythonentry point: transpose

def transpose(matrix):
    # O(m*n) time, O(m*n) space for the grid handed back: allocate the mirrored
    # shape once, then walk the readings in row order and drop each one straight
    # into its mirrored slot.
    rows = len(matrix)
    cols = len(matrix[0])
    out = [[0] * rows for _ in range(cols)]
    for r in range(rows):
        row = matrix[r]
        for c in range(cols):
            out[c][r] = row[c]
    return out

The obvious version

Build the answer one row at a time: for output row `c`, walk down every input row and pick out the reading in column `c`, gathering the column into a fresh list.

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_transpose(matrix):
    cols = len(matrix[0])
    result = []
    for c in range(cols):
        gathered = []
        for row in matrix:
            gathered.append(row[c])
        result.append(gathered)
    return result

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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