Big O Practice: Spiral Matrix

def spiral_order(matrix):
    if not matrix:
        return []
    m, n = len(matrix), len(matrix[0])
    res = []
    top, bottom = 0, m - 1
    left, right = 0, n - 1
    while top <= bottom and left <= right:
        for c in range(left, right + 1):
            res.append(matrix[top][c])
        top += 1
        for r in range(top, bottom + 1):
            res.append(matrix[r][right])
        right -= 1
        if top <= bottom:
            for c in range(right, left - 1, -1):
                res.append(matrix[bottom][c])
            bottom -= 1
        if left <= right:
            for r in range(bottom, top - 1, -1):
                res.append(matrix[r][left])
            left += 1
    return res

Variables

• m - number of rows in the matrix
• n - number of columns in the matrix

Explanation

Each cell is visited exactly once - **O(m · n)** time. Auxiliary space is **O(1)** if we exclude the output; counting the output makes it O(m · n).

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