054. Spiral Matrix

At a Glance

Topic: Array
Pattern: Analyze Pattern
Difficulty: Medium
LeetCode: 054

Problem Statement

Given an m x n matrix, return all elements of the matrix in spiral order.

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]

Constraints:

m == matrix.length
n == matrix[i].length
1 <= m, n <= 10
-100 <= matrix[i][j] <= 100

Approach & Solution Steps

Simulation with Visited Map — Use dx, dy arrays [0, 1, 0, -1]. If you hit an edge or a visited cell, turn 90 degrees. Time: $O(m \cdot n)$, Space: $O(m \cdot n)$ extra space for visited map. Suboptimal.
Layer-by-Layer Shrinking Boundaries — Track four integer walls. Time: $O(m \cdot n)$, Space: $O(1)$ extra space. (Always pick this)

Optimal JS Solution

function spiralOrder(matrix) {
	if (matrix.length === 0) return [];

	const result = [];
	let top = 0;
	let bottom = matrix.length - 1;
	let left = 0;
	let right = matrix[0].length - 1;

	while (top <= bottom && left <= right) {
		// 1. Top Row
		for (let c = left; c <= right; c++) {
			result.push(matrix[top][c]);
		}
		top++;

		// 2. Right Col
		for (let r = top; r <= bottom; r++) {
			result.push(matrix[r][right]);
		}
		right--;

		// Check condition before backwards sweeps
		if (top <= bottom && left <= right) {
			// 3. Bottom Row
			for (let c = right; c >= left; c--) {
				result.push(matrix[bottom][c]);
			}
			bottom--;

			// 4. Left Col
			for (let r = bottom; r >= top; r--) {
				result.push(matrix[r][left]);
			}
			left++;
		}
	}

	return result;
}

Edge Cases & Pitfalls

Always consider empty or null inputs.
Watch out for off-by-one index errors.