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Rotate Image

DESCRIPTION (credit Leetcode.com)

Write a function to rotate an n x n 2D matrix representing an image by 90 degrees clockwise. The rotation must be done in-place, meaning you should modify the input matrix directly without using an additional matrix for the operation.

EXAMPLES

Input:


matrix = [
    [1,4,7],
    [2,5,8],
    [3,6,9]
]

Output:


[
    [3,2,1],
    [6,5,4],
    [9,8,7]
]

Explanation: The matrix is rotated by 90 degrees clockwise, transforming its columns into rows in reverse order.

Run your code to see results here

Explanation

This problem can be done in two steps. We first transpose the matrix, then reverse the elements in each row.

Step 1:

Transpose the matrix by swapping the elements across the diagonal. This can be done in-place by using a nested for loop to swap the elements.


def rotate_image(matrix):
  n = len(matrix)

  # Transpose the matrix
  for i in range(n):
    for j in range(i, n):
      matrix[i][j], matrix[j][i] = \ 
        matrix[j][i], matrix[i][j]

  # Reverse each row
  for i in range(n):
    matrix[i] = matrix[i][::-1]
  return matrix

n = 3

0 / 12

Step 2:

Reverse the elements in each row of the matrix.


def rotate_image(matrix):
  n = len(matrix)

  # Transpose the matrix
  for i in range(n):
    for j in range(i, n):
      matrix[i][j], matrix[j][i] = \ 
        matrix[j][i], matrix[i][j]

  # Reverse each row
  for i in range(n):
    matrix[i] = matrix[i][::-1]
  return matrix

swap matrix[2][2] and matrix[2][2]

0 / 4

Solution


def rotate_image(matrix):
  n = len(matrix)

  # Transpose the matrix
  for i in range(n):
    for j in range(i, n):
      matrix[i][j], matrix[j][i] = \ 
        matrix[j][i], matrix[i][j]

  # Reverse each row
  for i in range(n):
    matrix[i] = matrix[i][::-1]
  return matrix

rotate image

0 / 17

Complexity Analysis

Time Complexity: O(n²) where n is the number of rows in the matrix. Space Complexity: O(1) since we are doing the rotation in-place.

Next: Set Matrix Zeroes

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DESCRIPTION (credit Leetcode.com)

EXAMPLES

Input:


matrix = [
    [1,4,7],
    [2,5,8],
    [3,6,9]
]

Output:


[
    [3,2,1],
    [6,5,4],
    [9,8,7]
]

Explanation: The matrix is rotated by 90 degrees clockwise, transforming its columns into rows in reverse order.

Run your code to see results here

Explanation

This problem can be done in two steps. We first transpose the matrix, then reverse the elements in each row.

Step 1:

Transpose the matrix by swapping the elements across the diagonal. This can be done in-place by using a nested for loop to swap the elements.


def rotate_image(matrix):
  n = len(matrix)

  # Transpose the matrix
  for i in range(n):
    for j in range(i, n):
      matrix[i][j], matrix[j][i] = \ 
        matrix[j][i], matrix[i][j]

  # Reverse each row
  for i in range(n):
    matrix[i] = matrix[i][::-1]
  return matrix

n = 3

0 / 12

Step 2:

Reverse the elements in each row of the matrix.


def rotate_image(matrix):
  n = len(matrix)

  # Transpose the matrix
  for i in range(n):
    for j in range(i, n):
      matrix[i][j], matrix[j][i] = \ 
        matrix[j][i], matrix[i][j]

  # Reverse each row
  for i in range(n):
    matrix[i] = matrix[i][::-1]
  return matrix

swap matrix[2][2] and matrix[2][2]

0 / 4

Solution


def rotate_image(matrix):
  n = len(matrix)

  # Transpose the matrix
  for i in range(n):
    for j in range(i, n):
      matrix[i][j], matrix[j][i] = \ 
        matrix[j][i], matrix[i][j]

  # Reverse each row
  for i in range(n):
    matrix[i] = matrix[i][::-1]
  return matrix

rotate image

0 / 17

Complexity Analysis

Time Complexity: O(n²) where n is the number of rows in the matrix. Space Complexity: O(1) since we are doing the rotation in-place.

Next: Set Matrix Zeroes

Your account is free and you can post anonymously if you choose.

Sort By

Loading comments...

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