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48. Spiral Matrix

mediumAsked at Vercel

Given a matrix, return all elements in spiral order. Vercel asks this to see if you can manage four mutually-shrinking bounds without off-by-one errors — a clean test of disciplined iteration.

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

Source citations

Public interview reports confirming this problem appears in Vercel loops.

  • Glassdoor (2025-12)— Vercel platform onsite; boundary-tracking expected.
  • Blind (2026-Q1)— Listed in Vercel screen pool.

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 + visited matrix

Walk in 4 directions; turn when you hit a wall 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;
  const 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: O(mn) extra space for visited. Boundary-shrinking version is cleaner.

2. Four-boundary shrinking (optimal)

Maintain top, bottom, left, right. Walk right (top++), down (right--), left (bottom--), up (left++). Stop when boundaries cross.

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

Tradeoff: O(1) extra space, four for-loops per layer. The `if` guards on the last two loops handle non-square shapes (e.g., 1xN) that would otherwise repeat rows/columns.

Vercel-specific tips

Vercel grades the boundary version. Bonus signal: the two `if` guards (must check top <= bot before walking left, and left <= right before walking up) — these are the classic off-by-one bugs that fail on 1xN or Nx1 matrices.

Common mistakes

  • Forgetting the `if` guards on the bottom/left passes — duplicate elements on single-row/column matrices.
  • Updating boundaries before the loop emits — off-by-one.
  • Using a visited matrix when boundary-tracking is cleaner.

Follow-up questions

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

  • Spiral Matrix II (LC 59) — fill an empty matrix with 1..n^2 in spiral.
  • Spiral Matrix III (LC 885) — start from arbitrary position.
  • Diagonal traversal (LC 498) — related coordinate-juggling.

Solve it now

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Output

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FAQ

Why the two extra `if` guards?

On a 1-row matrix, after walking right, top++ makes top > bot. Without the guard, the leftward walk would re-emit the same row. Same for 1-column shapes.

Could I do it recursively?

Yes — peel the outer layer and recurse on the inner submatrix. Conceptually clean, but boundary-tracking is more efficient (no recursion stack).

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