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Matrices

Rotate Image

medium

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.

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.

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You are given an `n x n` 2D matrix representing an image, return the marix such that image is rotated by 90 degrees (clockwise). For example, given the matrix: ``` [ [1,2,3], [4,5,6], [7,8,9] ] ``` The rotated matrix should be: ``` [ [7,4,1], [8,5,2], [9,6,3] ] ```

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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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Matrices

Rotate Image

DESCRIPTION (credit Leetcode.com)

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Explanation

Step 1:

Step 2:

Solution

Complexity Analysis

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