Spiral Matrix Medium 0 attempts
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Spiral Matrix

Medium ArrayTwo Pointers LeetCode

You're given a matrix with m rows and n columns. Traverse it in a spiral: start at the top-left cell and move right along the top row, drop down the rightmost column, sweep left along the bottom row, climb the leftmost column, then keep spiraling inward until every cell has been visited. Return the values in the order you visited them.

Edge cases: Matrices with just one row or one column. An empty matrix should return [].

Input Format
Implement spiralOrder(matrix). Inputs below show JavaScript function arguments; there is no stdin parsing.
Output Format
Return the result described in the statement.

Sample Input and Output

Example 1 · Run Sample

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]]
Output
[1,2,3]
Constraints
  • 0 <= matrix.length <= 10
  • All rows have the same length, at most 10.
  • -100 <= matrix[i][j] <= 100
Test Cases (2)
Case 1
Args: [[[1,2,3],[4,5,6],[7,8,9]]] Expected: [1,2,3,6,9,8,7,4,5]
Case 2
Args: [[[1,2,3]]] Expected: [1,2,3]
Topics
ArrayTwo Pointers

Approach: Layer-by-Layer Boundaries

Track four boundaries — top, bottom, left, right — and peel off one layer of the spiral per iteration. After each direction traversal, shrink the corresponding boundary and check whether you've crossed over.

javascript
function spiralOrder(matrix) {
  const result = [];
  if (!matrix.length) return result;
  let top = 0, bottom = matrix.length - 1;
  let left = 0, right = matrix[0].length - 1;

  while (top <= bottom && left <= right) {
    for (let i = left; i <= right; i++) result.push(matrix[top][i]);
    top++;
    for (let i = top; i <= bottom; i++) result.push(matrix[i][right]);
    right--;
    if (top <= bottom) {
      for (let i = right; i >= left; i--) result.push(matrix[bottom][i]);
      bottom--;
    }
    if (left <= right) {
      for (let i = bottom; i >= top; i--) result.push(matrix[i][left]);
      left++;
    }
  }
  return result;
}

Time Complexity: O(m × n) — every element is visited once

Space Complexity: O(1) extra space (excluding the output array)

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JavaScript