Microsoft · Coding 3 of 3

Spiral Matrix

Return the elements of a matrix in spiral order.

MediumMatrixSimulationBest: O(m·n) time · O(1) extra space

Practice question: an editorial practice prompt, not a claim that Microsoft asks this exact question. Interviews vary by role, level, team, and location.

Problem

Problem statement

Given an m × n matrix, return all of its elements in spiral order: along the top row left to right, down the right column, back along the bottom row, up the left column, then repeat on the next ring inward.

Examples

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]
Example 3
Input
matrix = [[1], [2], [3]]
Output
[1, 2, 3]

Why: A single column must not be read twice.

Constraints

  • 1 ≤ m, n ≤ 10
  • −100 ≤ matrix[i][j] ≤ 100
In the interview

How to answer it out loud

What it tests: Careful index management, loop invariants, and testing awkward shapes.

  1. 1
    Clarify

    Clarify the matrix shape and whether it is always square.

  2. 2
    Implement

    Keep four boundaries; read the top row, right column, bottom row and left column, then move the boundaries inward.

  3. 3
    Verify

    Test a single row, a single column, a 1 × 1 matrix, and a non-square matrix; explain O(m·n) time.

Approaches

From brute force to the best solution

ApproachIdeaTimeSpace
Direction simulationWalk and turn right whenever the next cell is outside the grid or already visited.O(m·n)O(m·n) for visited
Shrinking boundariesBestKeep top, bottom, left and right walls; read one ring, then move all four inward.O(m·n)O(1) extra
Key insight

The bugs live in the last ring. Only read the bottom row when it differs from the top row, and only read the left column when it differs from the right column.

Solution

Reference solution

Each version is self-contained and has been run against the examples and test cases on this page.

def spiral_order(matrix: list[list[int]]) -> list[int]:    result = []    top, bottom = 0, len(matrix) - 1    left, right = 0, len(matrix[0]) - 1    while top <= bottom and left <= right:        for c in range(left, right + 1):              # top row, left -> right            result.append(matrix[top][c])        for r in range(top + 1, bottom + 1):          # right column, downwards            result.append(matrix[r][right])        if top < bottom:                              # bottom row, if it is a new row            for c in range(right - 1, left - 1, -1):                result.append(matrix[bottom][c])        if left < right:                              # left column, if it is a new column            for r in range(bottom - 1, top, -1):                result.append(matrix[r][left])        top, bottom, left, right = top + 1, bottom - 1, left + 1, right - 1    return result
Time
O(m·n)
Space
O(1) extra
Approach
Shrinking boundaries
Test cases

Edge cases to run before you say “done”

InputExpectedWhy it matters
[[1, 2, 3]][1, 2, 3]A single row must not be read back.
[[1], [2], [3]][1, 2, 3]A single column.
[[7]][7]One cell.
Likely follow-ups

Answer the main question first, then take these on one at a time:

  • Can you fill an n × n matrix with 1 to n² in spiral order?
  • How would you walk the spiral counter-clockwise?
  • Can you do it with a direction array instead of four loops?
Avoid these mistakes

Check your recording or written solution against this list:

  • Reading the last row or column twice
  • Off-by-one boundaries on non-square matrices
  • Using a visited grid when boundaries are enough
Sources and methodology

This guide combines general interview-practice patterns with the public hiring material below. The practice prompt and coaching are PiriPiri AI editorial content, not official Microsoft questions or answers.

Microsoft Careers: technical interviewing ↗
Official sources reviewed 20 September 2026.