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50. Spiral Matrix

mediumAsked at Reddit

Traverse a matrix in spiral order. Reddit uses this to test boundary tracking — the same skill needed when iterating around 2D region tiles in their image-grid rendering for AR/AR filters.

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

Source citations

Public interview reports confirming this problem appears in Reddit loops.

  • Glassdoor (2026-Q1)— Reddit phone screen for backend roles.

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

Track (dr, dc) direction. Turn right when you hit a boundary or a visited cell.

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

Tradeoff: Clean, but requires O(m*n) visited array.

2. Four boundaries shrinking (optimal)

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

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

Tradeoff: O(1) extra. The two extra guards (if top <= bottom, if left <= right) handle rectangle-not-square cases.

Reddit-specific tips

Reddit interviewers want the four-boundary approach. Bonus signal: explicitly walk through a 3x4 to show why the bottom-row and left-column iterations need guard conditions.

Common mistakes

  • Forgetting the guards — duplicates rows/cols when matrix becomes a single row or column.
  • Off-by-one (use <= and -- after the loop, or < and don't change).
  • Using direction array without revisiting prevention.

Follow-up questions

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

  • Spiral Matrix II (LC 59) — fill a matrix in spiral.
  • Spiral Matrix III (LC 885) — start from arbitrary point.
  • Output diagonals instead of spiral.

Solve it now

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Output

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FAQ

Why need top <= bottom check before the third side?

After processing the top row in a single-row matrix, top > bottom; without the guard you'd re-output the only row in reverse.

Could we do this recursively?

Yes — process outer ring, recurse on (m-2 x n-2) inner. Same complexity, more memory.

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