Python3 Program to Inplace rotate square matrix by 90 degrees | Set 1

Given a square matrix, turn it by 90 degrees in anti-clockwise direction without using any extra space.
Examples :

Input:
Matrix:
 1  2  3
 4  5  6
 7  8  9
Output:
 3  6  9 
 2  5  8 
 1  4  7 
The given matrix is rotated by 90 degree 
in anti-clockwise direction.

Input:
 1  2  3  4 
 5  6  7  8 
 9 10 11 12 
13 14 15 16 
Output:
 4  8 12 16 
 3  7 11 15 
 2  6 10 14 
 1  5  9 13
The given matrix is rotated by 90 degree 
in anti-clockwise direction.

An approach that requires extra space is already discussed here.
Approach: To solve the question without any extra space, rotate the array in form of squares, dividing the matrix into squares or cycles. For example, 
A 4 X 4 matrix will have 2 cycles. The first cycle is formed by its 1st row, last column, last row and 1st column. The second cycle is formed by 2nd row, second-last column, second-last row and 2nd column. The idea is for each square cycle, swap the elements involved with the corresponding cell in the matrix in anti-clockwise direction i.e. from top to left, left to bottom, bottom to right and from right to top one at a time using nothing but a temporary variable to achieve this.
Demonstration:

First Cycle (Involves Red Elements)
 1  2  3 4 
 5  6  7 8 
 9 10 11 12 
 13 14 15 16 

Moving first group of four elements (First
elements of 1st row, last row, 1st column 
and last column) of first cycle in counter
clockwise. 
 4  2  3 16
 5  6  7 8 
 9 10 11 12 
 1 14  15 13 
 
Moving next group of four elements of 
first cycle in counter clockwise 
 4  8  3 16 
 5  6  7  15  
 2  10 11 12 
 1  14  9 13 

Moving final group of four elements of 
first cycle in counter clockwise 
 4  8 12 16 
 3  6  7 15 
 2 10 11 14 
 1  5  9 13 


Second Cycle (Involves Blue Elements)
 4  8 12 16 
 3  6 7  15 
 2  10 11 14 
 1  5  9 13 

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

Algorithm:

1. There is N/2 squares or cycles in a matrix of side N. Process a square one at a time. Run a loop to traverse the matrix a cycle at a time, i.e loop from 0 to N/2 – 1, loop counter is i
2. Consider elements in group of 4 in current square, rotate the 4 elements at a time. So the number of such groups in a cycle is N – 2*i.
3. So run a loop in each cycle from x to N – x – 1, loop counter is y
4. The elements in the current group is (x, y), (y, N-1-x), (N-1-x, N-1-y), (N-1-y, x), now rotate the these 4 elements, i.e (x, y) <- (y, N-1-x), (y, N-1-x)<- (N-1-x, N-1-y), (N-1-x, N-1-y)<- (N-1-y, x), (N-1-y, x)<- (x, y)
5. Print the matrix.

Python3

# Python3 program to rotate a matrix by 90 degrees
N = 4

# An Inplace function to rotate
# N x N matrix by 90 degrees in
# anti-clockwise direction
def rotateMatrix(mat):

    # Consider all squares one by one
    for x in range(0, int(N / 2)):

        # Consider elements in group
        # of 4 in current square
        for y in range(x, N-x-1):

            # store current cell in temp variable
            temp = mat[x][y]

            # move values from right to top
            mat[x][y] = mat[y][N-1-x]

            # move values from bottom to right
            mat[y][N-1-x] = mat[N-1-x][N-1-y]

            # move values from left to bottom
            mat[N-1-x][N-1-y] = mat[N-1-y][x]

            # assign temp to left
            mat[N-1-y][x] = temp


# Function to print the matrix
def displayMatrix( mat ):

    for i in range(0, N):

        for j in range(0, N):

            print (mat[i][j], end = ' ')
        print ("")




# Driver Code
mat = [[0 for x in range(N)] for y in range(N)]

# Test case 1
mat = [ [1, 2, 3, 4 ],
        [5, 6, 7, 8 ],
        [9, 10, 11, 12 ],
        [13, 14, 15, 16 ] ]

'''
# Test case 2
mat = [ [1, 2, 3 ],
        [4, 5, 6 ],
        [7, 8, 9 ] ]

# Test case 3
mat = [ [1, 2 ],
        [4, 5 ] ]

'''

rotateMatrix(mat)

# Print rotated matrix
displayMatrix(mat)


# This code is contributed by saloni1297

Output :

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

Complexity Analysis:

• Time Complexity: O(n*n), where n is size of array. 
  A single traversal of the matrix is needed.
• Space Complexity: O(1). 
  As a constant space is needed

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