How to use Transpose and Reverse In Javascript
This approach first transposes the matrix and then reverses each row to achieve the clockwise rotation.
Step 1: Transpose the Matrix: Swap matrix[i][j] with matrix[j][i].
Step 2: Reverse Each Row: Reverse each row of the transposed matrix.
Example: In this example The rotateMatrix function rotates a matrix clockwise by first transposing it (swapping rows and columns) and then reversing each row. It correctly rotates the given matrices in both test cases.
// Function to rotate a matrix clockwise
function rotateMatrix(matrix) {
const n = matrix.length;
// Transpose the matrix
for (let i = 0; i < n; i++) {
for (let j = i; j < n; j++) {
// Swap matrix[i][j] with matrix[j][i]
[matrix[i][j], matrix[j][i]] = [matrix[j][i], matrix[i][j]];
}
}
// Reverse each row
for (let i = 0; i < n; i++) {
matrix[i].reverse();
}
return matrix;
}
// Test Case 1
let matrix1 = [
[1, 2, 3, 4],
[5, 6, 7, 8],
[9, 10, 11, 12],
[13, 14, 15, 16]
];
console.log("Original Matrix:");
console.log(matrix1);
console.log("Rotated Matrix:");
console.log(rotateMatrix(matrix1));
// Test Case 2
let matrix2 = [
[1, 2, 3],
[4, 5, 6],
[7, 8, 9]
];
console.log("Original Matrix:");
console.log(matrix2);
console.log("Rotated Matrix:");
console.log(rotateMatrix(matrix2));
Output
Original Matrix: [ [ 1, 2, 3, 4 ], [ 5, 6, 7, 8 ], [ 9, 10, 11, 12 ], [ 13, 14, 15, 16 ] ] Rotated Matrix: [ [ 13, 9, 5, 1 ], [ 14, 10, 6, 2 ], [ 15, 11, 7, 3 ], [ 16, 12, 8, 4 ] ] Ori...
Time Complexity: O(N2)
Space Complexity: O(1)
Please refer complete article on Rotate Matrix Elements for more details!
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.
<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)