How do you rotate a two dimensional array?

Question

Inspired by Raymond Chen's post, say you have a 4x4 two dimensional array, write a function that rotates it 90 degrees. Raymond links to a solution in pseudo code, but I'd like to see some real world stuff.

[1][2][3][4]
[5][6][7][8]
[9][0][1][2]
[3][4][5][6]

Becomes:

[3][9][5][1]
[4][0][6][2]
[5][1][7][3]
[6][2][8][4]

Update: Nick's answer is the most straightforward, but is there a way to do it better than n^2? What if the matrix was 10000x10000?

Answer 1

This is my implementation, in C, O(1) memory complexity, in place rotation, 90 degrees clockwise:

#include <stdio.h>

#define M_SIZE 5

static void initMatrix();
static void printMatrix();
static void rotateMatrix();

static int m[M_SIZE][M_SIZE];

int main(void){
    initMatrix();
    printMatrix();
    rotateMatrix();
    printMatrix();

    return 0;
}

static void initMatrix(){
    int i, j;

    for(i = 0; i < M_SIZE; i++){
        for(j = 0; j < M_SIZE; j++){
            m[i][j] = M_SIZE*i + j + 1;
        }
    }
}

static void printMatrix(){
    int i, j;

    printf("Matrix\n");
    for(i = 0; i < M_SIZE; i++){
        for(j = 0; j < M_SIZE; j++){
            printf("%02d ", m[i][j]);
        }
        printf("\n");
    }
    printf("\n");
}

static void rotateMatrix(){
    int r, c;

    for(r = 0; r < M_SIZE/2; r++){
        for(c = r; c < M_SIZE - r - 1; c++){
            int tmp = m[r][c];

            m[r][c] = m[M_SIZE - c - 1][r];
            m[M_SIZE - c - 1][r] = m[M_SIZE - r - 1][M_SIZE - c - 1];
            m[M_SIZE - r - 1][M_SIZE - c - 1] = m[c][M_SIZE - r - 1];
            m[c][M_SIZE - r - 1] = tmp;
        }
    }
}

Answer 2

For i:= 0 to X do For j := 0 to X do graphic[j][i] := graphic2[X-i][j]

X is the size of the array the graphic is in.

Answer 3

short normal[4][4] = {{8,4,7,5},{3,4,5,7},{9,5,5,6},{3,3,3,3}};

short rotated[4][4];

for (int r = 0; r < 4; ++r)
{
  for (int c = 0; c < 4; ++c)
  {
    rotated[r][c] = normal[c][3-r];
  }
}

Simple C++ method, tho there would be a big memory overhead in a big array.

Answer 4

Implementation of dimple's +90 pseudocode (e.g. transpose then reverse each row) in JavaScript:

function rotate90(a){
  // transpose from http://www.codesuck.com/2012/02/transpose-javascript-array-in-one-line.html
  a = Object.keys(a[0]).map(function (c) { return a.map(function (r) { return r[c]; }); });
  // row reverse
  for (i in a){
    a[i] = a[i].reverse();
  }
  return a;
}

Answer 5

You can do this in 3 easy steps:

1)Suppose we have a matrix

   1 2 3
   4 5 6
   7 8 9

2)Take the transpose of the matrix

   1 4 7
   2 5 8
   3 6 9

3)Interchange rows to get rotated matrix

   3 6 9
   2 5 8
   1 4 7

Java source code for this:

public class MyClass {

    public static void main(String args[]) {
        Demo obj = new Demo();
        /*initial matrix to rotate*/
        int[][] matrix = { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 } };
        int[][] transpose = new int[3][3]; // matrix to store transpose

        obj.display(matrix);              // initial matrix

        obj.rotate(matrix, transpose);    // call rotate method
        System.out.println();
        obj.display(transpose);           // display the rotated matix
    }
}

class Demo {   
    public void rotate(int[][] mat, int[][] tran) {

        /* First take the transpose of the matrix */
        for (int i = 0; i < mat.length; i++) {
            for (int j = 0; j < mat.length; j++) {
                tran[i][j] = mat[j][i]; 
            }
        }

        /*
         * Interchange the rows of the transpose matrix to get rotated
         * matrix
         */
        for (int i = 0, j = tran.length - 1; i != j; i++, j--) {
            for (int k = 0; k < tran.length; k++) {
                swap(i, k, j, k, tran);
            }
        }
    }

    public void swap(int a, int b, int c, int d, int[][] arr) {
        int temp = arr[a][b];
        arr[a][b] = arr[c][d];
        arr[c][d] = temp;    
    }

    /* Method to display the matrix */
    public void display(int[][] arr) {
        for (int i = 0; i < arr.length; i++) {
            for (int j = 0; j < arr.length; j++) {
                System.out.print(arr[i][j] + " ");
            }
            System.out.println();
        }
    }
}

Output:

1 2 3 
4 5 6 
7 8 9 

3 6 9 
2 5 8 
1 4 7 

Answer 6

Whilst rotating the data in place might be necessary (perhaps to update the physically stored representation), it becomes simpler and possibly more performant to add a layer of indirection onto the array access, perhaps an interface:

interface IReadableMatrix
{
    int GetValue(int x, int y);
}

If your Matrix already implements this interface, then it can be rotated via a decorator class like this:

class RotatedMatrix : IReadableMatrix
{
    private readonly IReadableMatrix _baseMatrix;

    public RotatedMatrix(IReadableMatrix baseMatrix)
    {
        _baseMatrix = baseMatrix;
    }

    int GetValue(int x, int y)
    {
        // transpose x and y dimensions
        return _baseMatrix(y, x);
    }
}

Rotating +90/-90/180 degrees, flipping horizontally/vertically and scaling can all be achieved in this fashion as well.

Performance would need to be measured in your specific scenario. However the O(n^2) operation has now been replaced with an O(1) call. It's a virtual method call which is slower than direct array access, so it depends upon how frequently the rotated array is used after rotation. If it's used once, then this approach would definitely win. If it's rotated then used in a long-running system for days, then in-place rotation might perform better. It also depends whether you can accept the up-front cost.

As with all performance issues, measure, measure, measure!