Hackerrank - Matrix Layer Rotation Solution

You are given a 2D matrix of dimension  and a positive integer . You have to rotate the matrix  times and print the resultant matrix. Rotation should be in anti-clockwise direction.

Rotation of a  matrix is represented by the following figure. Note that in one rotation, you have to shift elements by one step only.

It is guaranteed that the minimum of m and n will be even.

As an example rotate the Start matrix by 2:

Start         First           Second
 1 2 3 4        2  3  4  5      3  4  5  6
12 1 2 5  ->   1  2  3  6 ->   2  3  4  7
11 4 3 6      12  1  4  7       1  2  1  8
10 9 8 7      11 10  9  8     12 11 10  9

Function Description

Complete the matrixRotation function in the editor below. It should print the resultant 2D integer array and return nothing.

matrixRotation has the following parameter(s):

• matrix: a 2D array of integers
• r: an integer that represents the rotation factor

Input Format

The first line contains three space separated integers, , , and , the number of rows and columns in , and the required rotation.
The next  lines contain  space-separated integers representing the elements of a row of .

Constraints

Output Format

Print each row of the rotated matrix as space-separated integers on separate lines.

Sample Input

Sample Input #01

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

Sample Output #01

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

Explanation #01

The matrix is rotated through two rotations.

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

Sample Input #02

5 4 7
1 2 3 4
7 8 9 10
13 14 15 16
19 20 21 22
25 26 27 28

Sample Output #02

28 27 26 25
22 9 15 19
16 8 21 13
10 14 20 7
4 3 2 1

Explanation 02

The various states through 7 rotations:

1  2  3  4      2  3  4 10    3  4 10 16    4 10 16 22
7  8  9 10      1  9 15 16    2 15 21 22    3 21 20 28
13 14 15 16 ->  7  8 21 22 -> 1  9 20 28 -> 2 15 14 27 ->
19 20 21 22    13 14 20 28    7  8 14 27    1  9  8 26
25 26 27 28    19 25 26 27    13 19 25 26   7 13 19 25

10 16 22 28    16 22 28 27    22 28 27 26    28 27 26 25
 4 20 14 27    10 14  8 26    16  8  9 25    22  9 15 19
 3 21  8 26 ->  4 20  9 25 -> 10 14 15 19 -> 16  8 21 13
 2 15  9 25     3 21 15 19     4 20 21 13    10 14 20  7
 1  7 13 19     2  1  7 13     3  2  1  7     4  3  2  1

Sample Input #03

2 2 3
1 1
1 1

Sample Output #03

1 1
1 1

Explanation #03

All of the elements are the same, so any rotation will repeat the same matrix.

Solution in Python

def matrixRotation(matrix,r):
    m, n = len(matrix), len(matrix[0])
    b = [[None]*n for _ in range(m)]
    indices = []

    for c in range(min(m,n)//2):
        index = []
        for i in range(c,n-c): index.append((c,i))
        for i in range(c+1,m-1-c): index.append((i,n-1-c))
        for i in range(c,n-c)[::-1]: index.append((m-1-c,i))
        for i in range(1+c,m-1-c)[::-1]: index.append((i,c))
        if not index:
            break
        indices.append(index)

    rotated = []
    for index in indices:
        k = r%len(index)
        rotated.append(index[k:]+index[:k])

    for (x,y) in zip(indices,rotated):
        for ((c,d),(e,f)) in zip(x,y):
            b[c][d] = matrix[e][f]

    return b
m,n,r = map(int,input().split())
matrix = [input().split() for i in range(m)]
for i in matrixRotation(matrix,r):
    print(*i)