The Spiral Harvest

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
Asked in
Citadel

Problem

Given a 2D matrix, return all elements in spiral order starting top-left, going clockwise (right, down, left, up), layer by layer.

The snail reads the grid in its own special order.

Example 1

Input:

matrix = [[1,2,3],[4,5,6],[7,8,9]]

Output:

[1,2,3,6,9,8,7,4,5]

Example 2

Input:

matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]

Output:

[1,2,3,4,8,12,11,10,9,5,6,7]

Worked solution and explanation

Why this problem exists in real interviews

This tests boundary tracking and direction cycling in a 2D matrix. Spiral traversal requires careful management of four boundaries (top, bottom, left, right) and is a common probe for attention to detail in index manipulation.

Break down the requirements

Step 1: Initialize four boundary pointers

Set top = 0, bottom = rows - 1, left = 0, right = cols - 1. These shrink inward as layers are consumed.

Step 2: Traverse right, then down, then left, then up

After completing each direction, shrink the corresponding boundary inward by one.

Step 3: Stop when boundaries cross

The loop ends when top > bottom or left > right.

The solution

Four-boundary shrinking traversal
def spiral_order(matrix: list) -> list:
    if not matrix or not matrix[0]:
        return []
    result = []
    top = 0
    bottom = len(matrix) - 1
    left = 0
    right = len(matrix[0]) - 1
    while top <= bottom and left <= right:
        for col in range(left, right + 1):
            result.append(matrix[top][col])
        top += 1
        for row in range(top, bottom + 1):
            result.append(matrix[row][right])
        right -= 1
        if top <= bottom:
            for col in range(right, left - 1, -1):
                result.append(matrix[bottom][col])
            bottom -= 1
        if left <= right:
            for row in range(bottom, top - 1, -1):
                result.append(matrix[row][left])
            left += 1
    return result

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