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Matrices

Spiral Matrix

medium

DESCRIPTION (credit Leetcode.com)

Write a function to traverse an m x n matrix in spiral order and return all elements in a single list. The traversal should start from the top left corner and proceed clockwise, spiraling inward until every element has been visited.

Input:


matrix = [
    [0,1,2],
    [3,4,5],
    [6,7,8]
]

Output:


[0,1,2,5,8,7,6,3,4]

Explanation: The elements of the matrix are returned in the order they are visited in a clockwise spiral starting from the top left corner.

Explanation

This solution uses 4 steps to traverse the matrix in spiral order:

1. Top Row

The first step is to pop the first row of the matrix, while copying those elements into the result array from left to right.


def spiral_order(matrix):
  result = []
  while matrix:
    result += matrix.pop(0)
    if matrix and matrix[0]:
      for row in matrix:
        result.append(row.pop())
    if matrix:
      result += matrix.pop()[::-1]
    if matrix and matrix[0]:
      for row in matrix[::-1]:
        result.append(row.pop(0))
  return result

initialize result

0 / 1

2. Right Column

Next, we traverse the right column of the matrix, popping the last element of each remaining row in matrix and copying those elements into the result array from top to bottom.


def spiral_order(matrix):
  result = []
  while matrix:
    result += matrix.pop(0)
    if matrix and matrix[0]:
      for row in matrix:
        result.append(row.pop())
    if matrix:
      result += matrix.pop()[::-1]
    if matrix and matrix[0]:
      for row in matrix[::-1]:
        result.append(row.pop(0))
  return result

traverse top row

0 / 2

3. Bottom Row

Next, we traverse the bottom row of the matrix, popping the last row of matrix and copying those elements into the result array from right to left.


def spiral_order(matrix):
  result = []
  while matrix:
    result += matrix.pop(0)
    if matrix and matrix[0]:
      for row in matrix:
        result.append(row.pop())
    if matrix:
      result += matrix.pop()[::-1]
    if matrix and matrix[0]:
      for row in matrix[::-1]:
        result.append(row.pop(0))
  return result

traverse right column

0 / 1

4. Left Column

Finally, we traverse the left column of the matrix, popping the first element of each remaining row in matrix and copying those elements into the result array from bottom to top.


def spiral_order(matrix):
  result = []
  while matrix:
    result += matrix.pop(0)
    if matrix and matrix[0]:
      for row in matrix:
        result.append(row.pop())
    if matrix:
      result += matrix.pop()[::-1]
    if matrix and matrix[0]:
      for row in matrix[::-1]:
        result.append(row.pop(0))
  return result

traverse bottom row

0 / 1

Repeat

At this point, we have traversed the perimeter of the original matrix in a clockwise spiral order. We repeat the process for the inner submatrix, until there are no more elements left to traverse.


def spiral_order(matrix):
  result = []
  while matrix:
    result += matrix.pop(0)
    if matrix and matrix[0]:
      for row in matrix:
        result.append(row.pop())
    if matrix:
      result += matrix.pop()[::-1]
    if matrix and matrix[0]:
      for row in matrix[::-1]:
        result.append(row.pop(0))
  return result

traverse left column

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Solution


def spiral_order(matrix):
  result = []
  while matrix:
    result += matrix.pop(0)
    if matrix and matrix[0]:
      for row in matrix:
        result.append(row.pop())
    if matrix:
      result += matrix.pop()[::-1]
    if matrix and matrix[0]:
      for row in matrix[::-1]:
        result.append(row.pop(0))
  return result

spiral matrix

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Complexity Analysis

Time complexity: O(m * n) where m is the number of rows and n is the number of columns (we visit each element in the matrix once).
Space complexity: O(m * n) for the result array.

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Matrices

Spiral Matrix

DESCRIPTION (credit Leetcode.com)

Explanation

1. Top Row

2. Right Column

3. Bottom Row

4. Left Column

Repeat

Solution

Complexity Analysis

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