Matrix Rotation - Hackerrank

https://www.hackerrank.com/challenges/matrix-rotation-algo

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

Rotation of a 4x5 matrix is represented by the following figure. Note that in one rotation, you have to shift elements by one step only (refer sample tests for more clarity).

It is guaranteed that the minimum of M and N will be even.

http://techcmxiv.blogspot.in/2015/12/matrix-rotation.html?view=magazine

Approach #2

For this successful try, I browsed the discussion section of problem and found out that I have to normalize the rotation for each of the rectangles. Check out the accepted code here.

This code was accepted because of its relatively low time complexity, let us assume that

Ri = # of rotations for ith rectangle of the matrix
Ri = R%(2*(M+N-4*i-2) where M = # of Rows, N = # of Cols & R = # of rotations.
Then, the run time would be (T0*R0 + T1*R1 + T2*R2 + .... + Tlimit * Rlimit)

Now,  Ri is calculated as R%(# of elements in the rectangle or the length of the rectangle)

https://gist.github.com/cmxiv/6ee0f6b37e4ab36ac947
//Rotate - Definition
void rotate() {
    long limit = M>N?N/2:M/2;
    for(long i=0; i<limit; i++) {
    for(long r=0; r<R%(2*(M+N-4*i-2)); r++)
    rotate(i);
    }
}

void rotate(long diag) {
    long col, row, temp = matrix[diag][diag];
    row = col = diag;


    for(; col<N-diag-1; col++) matrix[row][col] = matrix[row][col+1];
    for(; row<M-diag-1; row++) matrix[row][col] = matrix[row+1][col];
    for(; col>diag; col--) matrix[row][col] = matrix[row][col-1];
    for(; row>diag; row--) matrix[row][col] = matrix[row-1][col];

    matrix[diag+1][diag] = temp;
}
http://www.martinkysel.com/hackerrank-matrix-rotation-solution/

I first extract layers, to simplify the logic. Then, I rotate the layers similarly to the Codility Rotation challenge.

def reverse(arr, i, j):

    for idx in xrange((j - i + 1) / 2):

        arr[i + idx], arr[j - idx] = arr[j - idx], arr[i + idx]

def rotateList(A, K):

    # performs a right rotation

    # K needs to be adjusted to len(A) - K for left rotation

    l = len(A)

    K %= len(A)

    reverse(A, l - K, l - 1)

    reverse(A, 0, l - K - 1)

    reverse(A, 0, l - 1)

    return A

def rotateLayers(N, M, R, layers):

    for layer in layers:

        rotateList(layer, len(layer) - R)

def rotateMatrix(M, N, R, mat):

    # generate a list of layers

    l = int(min(N, M) // 2)

    layers = [[] for _ in range(l)]

    for level in xrange(l):

        top = (N - 1) - 2 * level

        side = (M - 1) - 2 * level

        for i in range(top):  # right

            layers[level].append(mat[level][level + i])

        for j in range(side):  # down

            layers[level].append(mat[level + j][level + top])

        for i in range(top):  # left

            layers[level].append(mat[level + side][level + top - i])

        for j in range(side):  # up

            layers[level].append(mat[level + side - j][level])

    # rotate each layer

    rotateLayers(N, M, R, layers)

    # fill the layers back in

    for level in range(l):

        top = (N - 1) - 2 * level

        side = (M - 1) - 2 * level

        for i in xrange(top):

            mat[level][level + i] = layers[level].pop(0)  # right

        for j in range(side):

            mat[level + j][level + top] = layers[level].pop(0)  # down

        for i in range(top):

            mat[level + side][level + top - i] = layers[level].pop(0)  # left

        for j in range(side):

            mat[level + side - j][level] = layers[level].pop(0)  # up

def main():

    M, N, R = map(int, raw_input().split())

    mat = []

    for i in range(M):

        mat.append(list(map(int, raw_input().split())))

    rotateMatrix(M, N, R, mat)

    # print the rotated matrix

    for row in range(M):

        for col in range(N):

            print mat[row][col],

        print

For the first try, I thought for writing the algorithm such that it would rotate the matrix once in anti-clockwise direction and then I would create a loop which would the run the said logic R times! This approach produces the correct outputs but the run time of the code is much much higher for large inputs. 

int main() {

    cin>>M>>N>>R;

    initMatrix();

    inputMatrix();

    for(long r=0; r<R; r++) rotate();

    outputMatrix();

    return 0;

}

//Matrix functions - Definition

void initMatrix() {

    matrix = new long* [M];

    for(long i=0; i<M; i++)

        matrix[i] = new long[N];

}

void inputMatrix() {

    for(long i=0; i<M; i++)

        for(long j=0; j<N; j++)

            cin>>matrix[i][j];

}

//Rotate - Definition

void rotate() {

    long limit = M>N?N/2:M/2;

    for(long i=0; i<limit; i++) rotate(i);

}

void rotate(long diag) {

    long col, row, temp = matrix[diag][diag];

    row = col = diag;

    

    for(; col<N-diag-1; col++) matrix[row][col] = matrix[row][col+1];

    for(; row<M-diag-1; row++) matrix[row][col] = matrix[row+1][col];

    for(; col>diag; col--) matrix[row][col] = matrix[row][col-1];

    for(; row>diag; row--) matrix[row][col] = matrix[row-1][col];

    matrix[diag+1][diag] = temp;

}

http://code.antonio081014.com/2011/11/interview-problem-rotate-matrix.html
 * Given an image represented by an NxN matrix, 
 * where each pixel in the image is 4 bytes, 
 * write a method to rotate the image by 90 degrees.
    public static void rotate(int[][] matrix, int n) {
        for (int i = 0; i < n / 2; i++) {
            int left = i;
            int right = n - 1 - i;
            for (int j = left; j < right; j++) {
                int offset = j - left;
                // save the top element;
                int tmp = matrix[left][left + offset];
                // right -> top;
                matrix[left][left + offset] = matrix[left + offset][right];
                // bottom -> right;
                matrix[left + offset][right] = matrix[right][right - offset];
                // left -> bottom;
                matrix[right][right - offset] = matrix[right - offset][left];
                // top -> left;
                matrix[right - offset][left] = tmp;
            }
        }
    }
https://github.com/careercup/CtCI-6th-Edition/blob/master/Java/Ch%2001.%20Arrays%20and%20Strings/Q1_07_Rotate_Matrix/Question.java

Rotate Matrix: Given an image represented by an NxN matrix, where each pixel in the image is 4 bytes, write a method to rotate the image by 90 degrees. Can you do this in place?

public static boolean rotate(int[][] matrix) {

if (matrix.length == 0 || matrix.length != matrix[0].length) return false; // Not a square

int n = matrix.length;

for (int layer = 0; layer < n / 2; layer++) {

int first = layer;

int last = n - 1 - layer;

for(int i = first; i < last; i++) {

int offset = i - first;

int top = matrix[first][i]; // save top

// left -> top

matrix[first][i] = matrix[last-offset][first];

// bottom -> left

matrix[last-offset][first] = matrix[last][last - offset]; 

// right -> bottom

matrix[last][last - offset] = matrix[i][last]; 

// top -> right

matrix[i][last] = top; // right <- saved top

}

}

return true;

}