Python Program for Rotate a Matrix by 180 degree

Given a square matrix, the task is that turn it by 180 degrees in an anti-clockwise direction without using any extra space. 

Examples :

Input:  1  2  3
         4  5  6
         7  8  9
         
Output: 9 8 7 
         6 5 4 
         3 2 1

Input:  1 2 3 4 
         5 6 7 8 
         9 0 1 2 
         3 4 5 6 
         
Output: 6 5 4 3 
         2 1 0 9 
         8 7 6 5 
         4 3 2 1

Method: 1
The solution of this problem is that to rotate a matrix by 180 degrees we can easily follow that step 

Matrix =  a00 a01 a02
          a10 a11 a12
          a20 a21 a22

when we rotate it by 90 degree
then matrix is
Matrix = a02 a12 a22
         a01 a11 a21
         a00 a10 a20
  
when we rotate it by again 90 
degree then the matrix is 
Matrix = a22 a21 a20
         a12 a11 a10
         a02 a01 a00 

From the above illustration, we get that simply to rotate the matrix by 180 degrees then we will have to print the given matrix in a reverse manner.

Python3

# Python3 program to
# rotate a matrix by
# 180 degrees
N = 3

# Function to Rotate
# the matrix by 180 degree


def rotateMatrix(mat):

    # Simply print from
    # last cell to first cell.
    i = N - 1
    while(i >= 0):
        j = N - 1
        while(j >= 0):
            print(mat[i][j], end=" ")
            j = j - 1
        print()
        i = i - 1


# Driven code
mat = [[1, 2, 3],
       [4, 5, 6],
       [7, 8, 9]]
rotateMatrix(mat)

Output

9 8 7 
6 5 4 
3 2 1 

Time complexity: O(N*N) 
Auxiliary Space: O(1)

Method : 2(In-place rotation)
There are four steps : 
1- Find transpose of a matrix. 
2- Reverse columns of the transpose. 
3- Find transpose of a matrix. 
4- Reverse columns of the transpose

Let the given matrix be
1  2  3  4
5  6  7  8
9  10 11 12
13 14 15 16

First we find transpose.
1 5 9 13
2 6 10 14
3 7 11 15
4 8 12 16

Then we reverse elements of every column.
4 8 12 16
3 7 11 15
2 6 10 14
1 5  9 13

then transpose again 
4 3 2 1 
8 7 6 5 
12 11 10 9
16 15 14 13 

Then we reverse elements of every column again
16 15 14 13 
12 11 10 9 
8 7 6 5 
4 3 2 1

Python3

# Python3 program for left rotation of matrix by 180

R = 4
C = 4

# Function to rotate the matrix by 180 degree


def reverseColumns(arr):
    for i in range(C):
        j = 0
        k = C-1
        while j < k:
            t = arr[j][i]
            arr[j][i] = arr[k][i]
            arr[k][i] = t
            j += 1
            k -= 1

# Function for transpose of matrix


def transpose(arr):
    for i in range(R):
        for j in range(i, C):
            t = arr[i][j]
            arr[i][j] = arr[j][i]
            arr[j][i] = t

# Function for display the matrix


def printMatrix(arr):
    for i in range(R):
        for j in range(C):
            print(arr[i][j], end=" ")
        print()

# Function to anticlockwise rotate matrix
# by 180 degree


def rotate180(arr):
    transpose(arr)
    reverseColumns(arr)
    transpose(arr)
    reverseColumns(arr)


# Driven code
arr = [[1, 2, 3, 4],
       [5, 6, 7, 8],
       [9, 10, 11, 12],
       [13, 14, 15, 16]]
rotate180(arr)
printMatrix(arr)

Output

16 15 14 13 
12 11 10 9 
8 7 6 5 
4 3 2 1 

Time complexity : O(R*C) 
Auxiliary Space : O(1)
In the code above, the transpose of the matrix has to be found twice, and also, columns have to be reversed twice. 
So, we can have a better solution.

Method : 3 (Position swapping)
Here, we swap the values in the respective positions. 

Python3

# Reverse Row at specified index in the matrix
def reverseRow(data, index):

    cols = len(data[index])
    for i in range(cols // 2):
        temp = data[index][i]
        data[index][i] = data[index][cols - i - 1]
        data[index][cols - i - 1] = temp

    return data

# Print Matrix data


def printMatrix(data):

    for i in range(len(data)):
        for j in range(len(data[0])):
            print(data[i][j], end=' ')

        print()

# Rotate Matrix by 180 degrees


def rotateMatrix(data):

    rows = len(data)
    cols = len(data[0])

    if (rows % 2):

        # If N is odd reverse the middle
        # row in the matrix
        data = reverseRow(data, len(data) // 2)

        # Swap the value of matrix [i][j] with
        # [rows - i - 1][cols - j - 1] for half
        # the rows size.
        for i in range(rows // 2):
            for j in range(cols):
                temp = data[i][j]
                data[i][j] = data[rows - i - 1][cols - j - 1]
                data[rows - i - 1][cols - j - 1] = temp

        return data


# Driver Code
data = [[1, 2, 3, 4, 5],
        [6, 7, 8, 9, 10],
        [11, 12, 13, 14, 15],
        [16, 17, 18, 19, 20],
        [21, 22, 23, 24, 25]]

# Rotate Matrix
data = rotateMatrix(data)

# Print Matrix
printMatrix(data)

Output

25 24 23 22 21 
20 19 18 17 16 
15 14 13 12 11 
10 9 8 7 6 
5 4 3 2 1 

Time complexity : O(R*C) 
Auxiliary Space : O(1)

Method #4: Reversing all rows using slicing

Python3

# Rotate Matrix by 180 degrees
def rotateMatrix(data):

    rows = len(data)
    cols = len(data[0])
    # Reversing all rows
    for i in range(len(data)):
        data[i] = data[i][::-1]
    # Reversing all rows of the matrix
    data = data[::-1]
    for i in range(rows):
        for j in range(cols):
            print(data[i][j], end=' ')
        print()


# Driver Code
data = [[1, 2, 3, 4, 5],
        [6, 7, 8, 9, 10],
        [11, 12, 13, 14, 15],
        [16, 17, 18, 19, 20],
        [21, 22, 23, 24, 25]]

# Rotate Matrix
rotateMatrix(data)

Output

25 24 23 22 21 
20 19 18 17 16 
15 14 13 12 11 
10 9 8 7 6 
5 4 3 2 1 

Please refer complete article on Rotate a Matrix by 180 degree for more details!