Skip to main content

50. Spiral Matrix

mediumAsked at Snowflake

Traverse an m x n matrix in spiral order. Snowflake asks this to test boundary tracking — relevant for windowing operations over multi-dimensional storage layouts.

By Alex Chen, Founder, InterviewChamp.AI · Last verified

Source citations

Public interview reports confirming this problem appears in Snowflake loops.

  • Glassdoor (2025-Q4)— Snowflake new-grad onsite as a boundary-tracking test.
  • LeetCode Discuss (2025-09)— Reported at Snowflake SDE-I screens.

Problem

Given an m x n matrix, return all elements of the matrix in spiral order.

Constraints

  • m == matrix.length
  • n == matrix[i].length
  • 1 <= m, n <= 10
  • -100 <= matrix[i][j] <= 100

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]

Approaches

1. Direction array with visited set

Walk in directions (right, down, left, up). Rotate direction on bounds or visited cell.

Time
O(mn)
Space
O(mn) for visited
function spiralOrder(matrix) {
  const m = matrix.length, n = matrix[0].length;
  const result = [];
  const seen = new Array(m).fill(null).map(() => new Array(n).fill(false));
  const dirs = [[0,1],[1,0],[0,-1],[-1,0]];
  let r = 0, c = 0, d = 0;
  for (let i = 0; i < m * n; i++) {
    result.push(matrix[r][c]);
    seen[r][c] = true;
    const nr = r + dirs[d][0], nc = c + dirs[d][1];
    if (nr < 0 || nr >= m || nc < 0 || nc >= n || seen[nr][nc]) {
      d = (d + 1) % 4;
    }
    r += dirs[d][0];
    c += dirs[d][1];
  }
  return result;
}

Tradeoff: O(mn) extra space for the seen matrix.

2. Four shrinking boundaries (optimal)

Track top, bottom, left, right. Walk right along top, down along right, left along bottom (if bottom > top), up along left (if right > left). Shrink boundaries after each side.

Time
O(mn)
Space
O(1)
function spiralOrder(matrix) {
  const result = [];
  let top = 0, bottom = matrix.length - 1, left = 0, right = matrix[0].length - 1;
  while (top <= bottom && left <= right) {
    for (let c = left; c <= right; c++) result.push(matrix[top][c]);
    top++;
    for (let r = top; r <= bottom; r++) result.push(matrix[r][right]);
    right--;
    if (top <= bottom) {
      for (let c = right; c >= left; c--) result.push(matrix[bottom][c]);
      bottom--;
    }
    if (left <= right) {
      for (let r = bottom; r >= top; r--) result.push(matrix[r][left]);
      left--;
    }
  }
  return result;
}

Tradeoff: O(1) extra space. Boundaries do all the work — no visited matrix.

Snowflake-specific tips

Snowflake interviewers want the boundary approach with the two if-guards for the bottom and left rows. Bonus signal: connect to 2D windowing — how would you traverse a 2D-tiled column store in row-major then column-major? The boundary tracking is the same pattern.

Common mistakes

  • Forgetting the top <= bottom guard before the bottom row — emits duplicates on single-row matrices.
  • Forgetting the left <= right guard before the left column — emits duplicates on single-column matrices.
  • Off-by-one when shrinking boundaries (decrementing before the loop runs).

Follow-up questions

An interviewer at Snowflake may pivot to one of these next:

  • Spiral Matrix II (LC 59) — generate, not traverse.
  • Walk in a snake/zigzag pattern.
  • How does Snowflake traverse 2D column tiles?

Solve it now

Free. No sign-up. Python and JavaScript run instantly in your browser.

Output

Press Run or Cmd+Enter to execute

FAQ

Why the bottom/left guards?

After top++ and right--, top might exceed bottom (single row exhausted) or right might exceed left. The guards prevent re-emitting already-visited cells.

Why not use a visited matrix?

It works but uses O(mn) extra space. Boundaries achieve O(1) extra space — important if the matrix is huge.

Practice these live with InterviewChamp.AI

Drill Spiral Matrix and other Snowflake interview questions under real-loop conditions with instant feedback on your reasoning, complexity claims, and code.

Practice these live with InterviewChamp.AI →