Javascript Program to Rotate Matrix Elements

Given a matrix, clockwise rotate elements in it.

Examples:

Input
1    2    3
4    5    6
7    8    9

Output:
4    1    2
7    5    3
8    9    6

For 4*4 matrix
Input:
1    2    3    4    
5    6    7    8
9    10   11   12
13   14   15   16

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

The idea is to use loops similar to the program for printing a matrix in spiral form. One by one rotate all rings of elements, starting from the outermost. To rotate a ring, we need to do following. 
    1) Move elements of top row. 
    2) Move elements of last column. 
    3) Move elements of bottom row. 
    4) Move elements of first column. 
Repeat above steps for inner ring while there is an inner ring.

Below is the implementation of above idea. Thanks to Gaurav Ahirwar for suggesting below solution. 

Javascript

<script>

// Javascript program to rotate a matrix

let R = 4;
let C = 4;

// A function to rotate a matrix
// mat[][] of size R x C.
// Initially, m = R and n = C
function rotatematrix(m, n, mat)
{
    let row = 0, col = 0;
    let prev, curr;

    /*
    row - Starting row index
    m - ending row index
    col - starting column index
    n - ending column index
    i - iterator
    */
    while (row < m && col < n)
    {
        if (row + 1 == m || col + 1 == n)
            break;

        // Store the first element of next
        // row, this element will replace
        // first element of current row
        prev = mat[row + 1][col];

        // Move elements of first row
        // from the remaining rows
        for(let i = col; i < n; i++)
        {
            curr = mat[row][i];
            mat[row][i] = prev;
            prev = curr;
        }
        row++;

        // Move elements of last column
        // from the remaining columns
        for(let i = row; i < m; i++)
        {
            curr = mat[i][n - 1];
            mat[i][n - 1] = prev;
            prev = curr;
        }
        n--;

        // Move elements of last row
        // from the remaining rows
        if (row < m)
        {
            for(let i = n - 1; i >= col; i--)
            {
                curr = mat[m - 1][i];
                mat[m - 1][i] = prev;
                prev = curr;
            }
        }
        m--;

        // Move elements of first column
        // from the remaining rows
        if (col < n)
        {
            for(let i = m - 1; i >= row; i--)
            {
                curr = mat[i][col];
                mat[i][col] = prev;
                prev = curr;
            }
        }
        col++;
    }

    // Print rotated matrix
    for(let i = 0; i < R; i++)
    {
        for(let j = 0; j < C; j++)
            document.write( mat[i][j] + " ");

        document.write("<br>");
    }
}

// Driver code

// Test Case 1
let a = [ [ 1, 2, 3, 4 ],
          [ 5, 6, 7, 8 ],
          [ 9, 10, 11, 12 ],
          [ 13, 14, 15, 16 ] ];

rotatematrix(R, C, a);

// This code is contributed by avanitrachhadiya2155

</script>

Output:

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

Time Complexity: O(max(m,n) * max(m,n))
Auxiliary Space: O(m*n)

Please refer complete article on Rotate Matrix Elements for more details!