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54. Spiral Matrix

mediumAsked at Goldman Sachs

Return all elements of a matrix in spiral order. Goldman Sachs uses Spiral Matrix to test boundary-tracking discipline — the candidate who articulates the four shrinking boundaries before coding is the one who survives the edge cases.

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

Source citations

Public interview reports confirming this problem appears in Goldman Sachs loops.

  • Glassdoor (2026-Q1)— Goldman Sachs SWE candidate reports list Spiral Matrix as a 'matrix-traversal boundary-tracking' question.
  • LeetCode Discuss (2025-12)— Spiral Matrix is in the top-20 of LeetCode's Goldman Sachs company tag.

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. Four shrinking boundaries (optimal)

Maintain top, bottom, left, right boundaries. Traverse top row, right col, bottom row, left col; shrink each boundary after.

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

Tradeoff: Linear in matrix size, O(1) extra space beyond the output. The two if-guards after the third and fourth traversals are critical — for non-square matrices, one dimension exhausts before the other and you'd double-visit cells without them.

2. Direction-vector with visited marker

Walk with a (dr, dc) vector; rotate clockwise when you hit a boundary or visited cell.

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

Tradeoff: More general framework but uses O(m*n) extra space for the visited grid. Mention as the 'reusable template for any spiral-walk problem', but lead with the boundaries version since Goldman wants O(1) extra space.

Goldman Sachs-specific tips

Goldman Sachs interviewers want to see the four boundaries named explicitly BEFORE coding. The exact edge case they probe: rectangular matrices where m != n. Single-row [1,2,3] should not loop back; single-column [[1],[2],[3]] should not skip cells. The two if-checks after the third and fourth traversals are the rubric point — without them, an mxn matrix where m != n returns duplicate cells.

Common mistakes

  • Forgetting the if (top <= bottom) and if (left <= right) guards for rectangular matrices.
  • Indexing matrix[top][i] when you meant matrix[i][right] — confusing row/col with the loop variable.
  • Hard-coding the assumption m == n, which silently breaks on non-square inputs.

Follow-up questions

An interviewer at Goldman Sachs may pivot to one of these next:

  • Spiral Matrix II (LC #59) — fill an n x n matrix with 1..n*n in spiral order. (Same template, fill instead of read.)
  • Spiral Matrix III (LC #885) — walk a spiral starting from an arbitrary cell, including out-of-grid positions.
  • What if you needed to traverse counter-clockwise? (Reverse the order of the four edge-traversals.)

Solve it now

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Output

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FAQ

Why do we need the two extra if-guards?

Because for a rectangular matrix, one dimension exhausts before the other. After traversing the top row of a 1xN matrix, top exceeds bottom and the third traversal (right-to-left bottom row) would re-visit cells already added. The guards skip the redundant traversal.

Is the boundary version really O(1) space?

Excluding the output. The result array itself is O(m*n) which is unavoidable since we're returning every cell. The space for state variables (top, bottom, left, right) is O(1).

Free learning resources

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