Against the Grain

A medium Python interview practice problem on DataDriven. Write the Python and run it against test cases, with instant feedback.

Domain
Python
Difficulty
medium
Seniority
L4

Problem

A metrics export comes back as a grid of equal-length rows, one row per day and one column per metric. Reorient it so each metric's column becomes its own row, and return an empty list when grid has no rows or its rows are empty.

Turn the grid on its side and let every column stand up as a row.

Example 1

Input:

grid = [[1,2,3],[4,5,6]]

Output:

[[1,4],[2,5],[3,6]]

Example 2

Input:

grid = [[1,2],[3,4],[5,6]]

Output:

[[1,3,5],[2,4,6]]

Worked solution and explanation

What this really is

Strip the spreadsheet costume and this is a coordinate swap: the cell sitting at row i, column j has to land at row j, column i. The whole problem turns on one realization that trips people up: the output's row index comes from the input's *column* index. Candidates who try to walk the input row by row end up either overwriting cells or building the wrong shape. The ones who get it iterate over columns on the outside and rows on the inside, and the code falls out cleanly.

Working through it

Step 1: Get the dimensions straight first

An m x n input (m days, n metrics) becomes an n x m output (n metrics, m days). Pin those dimensions down before writing a single loop, because that is the contract the result has to satisfy.

Step 2: Build one output row per input column

Drive the outer loop over columns of the input. Each column is one metric's run of values across every day, and that is exactly one row of the output. Reading a fixed column down every input row is the move that does the swap.

Step 3: Handle the empty shapes

Before indexing grid[0], guard the empty cases. An empty grid and a grid of empty rows both have nothing to reorient, so both return an empty list.

The solution

Outer loop over columns
def transpose(grid: list) -> list:
    if not grid or not grid[0]:
        return []
    rows = len(grid)
    cols = len(grid[0])
    result = []
    for c in range(cols):
        new_row = []
        for r in range(rows):
            new_row.append(grid[r][c])
        result.append(new_row)
    return result

The outer index runs over input columns, so each pass builds exactly one output row.

Loop over rows (wrong instinct)
Walking the input row by row appends whole input rows back, leaving the orientation unchanged or the indices tangled.
Loop over columns (correct)
Walking column by column collects each metric's values down the grid, producing one clean output row per metric.

Common follow-up questions