Spiral Matrix

Medium

Four shrinking boundaries

In simple words

Read the grid in a spiral: right across the top, down the right, left along the bottom, up the left, then inward.

The idea

Keep top, bottom, left, right borders. Walk right along top, down the right, left along bottom, up the left — shrinking the matching border after each pass until they cross.

←→stepspaceplay

1/18

Optimal

timeO(rows·cols)spaceO(1)

Peel the matrix ring by ring.

1while (top <= bottom && left <= right) {2  for (let c=left; c<=right; c++) out.push(M[top][c]);    top++;3  for (let r=top; r<=bottom; r++) out.push(M[r][right]);  right--;4  if (top<=bottom) for (let c=right;c>=left;c--) out.push(M[bottom][c]); bottom--;5  if (left<=right) for (let r=bottom;r>=top;r--) out.push(M[r][left]);   left++;6}

Input

grid: 4 × 4

Memory

at: —
cells marked: 0

Output

output: —
done: —

Check yourself

3 quick questions about this walkthrough. A wrong answer costs nothing.

Examples

Example 1

Input:: matrix = [[1,2,3],[4,5,6],[7,8,9]]
Output:: [1, 2, 3, 6, 9, 8, 7, 4, 5]
Explanation:: Peel the outer ring, then spiral inward.

Example 2

Input:: matrix = [[1,2],[3,4]]
Output:: [1, 2, 4, 3]
Explanation:: 1,2,4,3 winds around.

Example 3

Input:: matrix = [[1]]
Output:: [1]
Explanation:: A single cell.

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