Spiral Matrix

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Given an m × n matrix, return all of its elements in spiral order — starting at the top-left, moving right, then spiraling inward.

Example

Key Pointers

• Track four shrinking boundaries: top, bottom, left, right.
• Walk right across the top row, down the right column, left across the bottom row, up the left column — then shrink every boundary inward and repeat.
• Guard the last two directions each pass (bottom row, left column) with a boundary check, or a single remaining row/column gets walked twice.

Step-by-Step Analysis

1. Initialize top = 0, bottom = rows - 1, left = 0, right = cols - 1.
2. While top <= bottom and left <= right: walk left→right along row top, then increment top.
3. Walk top→bottom along column right, then decrement right.
4. If top <= bottom, walk right→left along row bottom, then decrement bottom.
5. If left <= right, walk bottom→top along column left, then increment left. Repeat until the boundaries cross.

Complexity

Time: O(m · n) — Space: O(1) extra, excluding the output

Code

def spiralOrder(matrix: list[list[int]]) -> list[int]:
    result = []
    top, bottom = 0, len(matrix) - 1
    left, right = 0, len(matrix[0]) - 1

    while top <= bottom and left <= right:
        for c in range(left, right + 1):
            result.append(matrix[top][c])
        top += 1

        for r in range(top, bottom + 1):
            result.append(matrix[r][right])
        right -= 1

        if top <= bottom:
            for c in range(right, left - 1, -1):
                result.append(matrix[bottom][c])
            bottom -= 1

        if left <= right:
            for r in range(bottom, top - 1, -1):
                result.append(matrix[r][left])
            left += 1

    return result

Code Explanation

How the solution above works, piece by piece.

1. Keep four boundaries: top, bottom, left, right.
2. Walk the top row left → right, then move top down. Walk the right column top → bottom, then move right left.
3. Walk the bottom row right → left and the left column bottom → top, but only if rows/columns remain (top <= bottom, left <= right). Without these checks a single leftover row or column would be read twice.
4. Repeat while the boundaries haven't crossed.
5. Every cell is visited once: O(m × n).

Visual Walkthrough

Running it on matrix = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]:

Video Explanation

NeetCode — Spiral Matrix, LeetCode 54