Algorithm of the Day (40 Part Series)

Today's algorithm is the Rotate Image problem:

You are given an n x n 2D matrix representing an image. Rotate the image by 90 degrees (clockwise).

You have to rotate the image in-place, which means you have to modify the input 2D matrix directly. DO NOT allocate another 2D matrix and do the rotation.

For example, if you were given the 2D array

```
[
[1,2,3],
[4,5,6],
[7,8,9]
]
```

rotating the array 90 degrees clockwise would give us the output of

```
[
[7,4,1],
[8,5,2],
[9,6,3]
]
```

Put another way, the first row becomes the last column, the second row becomes the middle column, and the last row becomes the first column.

![Three sets of images. In the first set, there's a 2D array, [[1,2,3], [4,5,6],[7,8,9]], and the first row is highlighted in cyan. There's a blue arrow turned 90 degrees clockwise, and next to it is another 2D array, [[, 1], [, 2],[, , 3]]. The last column is highlighted in cyan. In the second set, there's a 2D array, [[1,2,3], [4,5,6],[7,8,9]], whose second row is highlighted in cyan. There's a blue arrow turned 90 degrees clockwise, and next to it is another 2D array, [[, 4, 1], [, 5, 2], [, 6, 3]], whose second column is highlighted in cyan. In the third set of images, there's a 2D array, [[1,2,3], [4,5,6], [7,8,9]], and the last row is highlighted in cyan. There's a blue arrow turned 90 degrees clockwise, and next to it is another 2D array, [[7, 4, 1], [8, 5, 2], [9, 6, 3]], whose second column is highlighted in cyan. ](https://dev-to-uploads.s3.amazonaws.com/i/bluo1pumyica1dmly0qz.png)

In this post, I'll start by discussing my approach to solving this problem, then I'll code the solution using JavaScript.

## Approaching the Rotating 2D Array Problem

Not long ago, I discussed the problem of rotating a one dimensional array (you can find that post here). What's trickier about a 2D array is that you have to keep track of both the row and the column that we're in.

The way I'll be rotating the 2d array (also known as a matrix) is with a two step approach. First, I'll transpose the matrix, which means switching the rows with the columns. Then, I'll reverse the elements in each row.

Let's say our inputted matrix was

```
[
[1, 2],
[3, 4]
]
```

After transposing the matrix, it would look like this:

```
[
[1, 3],
[2, 4]
]
```

The first row became the first column, and the second row became the second column. However, we want all of these elements to be reversed, so we'll reverse each element in each row, giving us the final matrix:

```
[
[3, 1],
[4, 2]
]
```

which is the solution we're after.

## Solving the Matrix Rotation Problem

We'll start our solution by checking for edge cases. If the matrix is empty, then there's nothing to rotate, so we can immediately return `null`

. Additionally, because we know the matrix is square (n x n), if it has a length of 1, then it only has one element in it, so we can just return that element.

```
function rotate(matrix) {
if (!matrix.length) return null;
if (matrix.length === 1) return matrix;
//...
}
```

Now, like discussed above, we'll have a two step solution. To keep the code as neat as possible, we'll separate the steps out from the original `rotate`

function. We can create a separate function called `transpose()`

, which will take in the matrix, and we'll call it from inside the `rotate()`

function.

```
function rotate(matrix) {
if (!matrix.length) return null;
if (matrix.length === 1) return matrix;
transpose(matrix);
//...
}
function transpose(matrix) {
//...
}
```

Transposing the matrix, or switching the rows and columns, will require nested for loops. The first loop will go through each row, and the second loop will go through each column. Since they're nested, we'll be able to access each element at any row, column point. We'll start the first for loop at `i = 0`

, which is the first row, and we'll start the second for loop at `j = 1`

, which is the second column.

```
function rotate(matrix) {
if (!matrix.length) return null;
if (matrix.length === 1) return matrix;
transpose(matrix);
//...
}
function transpose(matrix) {
for (let i = 0; i < matrix.length; i++) {
for (let j = i; j < matrix[0].length; j++) {
//...
}
}
//...
}
```

Inside the for loops, we'll want to swap two elements -- the value at `matrix[i][j]`

will be swapped with the value at `matrix[j][i]`

. To do a swap, we need a temporary variable, called `temp`

, which enables us to store the value at one point before changing that point's value.

When the for loops are done executing, we can return the updated matrix back to `rotate()`

.

```
function rotate(matrix) {
if (!matrix.length) return null;
if (matrix.length === 1) return matrix;
transpose(matrix);
//...
}
function transpose(matrix) {
for (let i = 0; i < matrix.length; i++) {
for (let j = i; j < matrix[0].length; j++) {
const temp = matrix[i][j];
matrix[i][j] = matrix[j][i];
matrix[j][i] = temp;
}
}
return matrix;
}
```

We're now done with transposing the elements, so we have to move onto the second step of this solution: reversing the elements of each row. To do this, we'll want to go through each row in `matrix`

, and call a new function called `reverse()`

on that row. `reverse()`

will take in three arguments: the row we want to reverse, the starting point to reverse at (which is 0), and the ending point of the reversal (with is `row.length - 1`

).

```
function rotate(matrix) {
if (!matrix.length) return null;
if (matrix.length === 1) return matrix;
transpose(matrix);
matrix.forEach((row) => {
reverse(row, 0, row.length - 1);
});
}
function transpose(matrix) {
for (let i = 0; i < matrix.length; i++) {
for (let j = i; j < matrix[0].length; j++) {
const temp = matrix[i][j];
matrix[i][j] = matrix[j][i];
matrix[j][i] = temp;
}
}
return matrix;
}
function reverse(row, start, end) {
//...
}
```

Now, in `reverse()`

, we'll set up a while loop. The idea behind this function is to have two pointers, `start`

and `end`

. As long as the end pointer is larger than the start pointer, we'll want to swap the values at those two spots.

To start, therefore, we'll set up a while loop in `reverse(), which will keep going as long as`

start < end`.

``javascript`

function rotate(matrix) {

if (!matrix.length) return null;

if (matrix.length === 1) return matrix;

transpose(matrix);

matrix.forEach((row) => {

reverse(row, 0, row.length - 1);

});

}

function transpose(matrix) {

for (let i = 0; i < matrix.length; i++) {

for (let j = i; j < matrix[0].length; j++) {

const temp = matrix[i][j];

matrix[i][j] = matrix[j][i];

matrix[j][i] = temp;

}

}

return matrix;

}

function reverse(row, start, end) {

while (start < end) {

//...

}

//...

}

`

Just like we did in `transpose()`

, we'll need to set up a temporary variable in order to swap the values at the start and end points.

``javascript`

function rotate(matrix) {

if (!matrix.length) return null;

if (matrix.length === 1) return matrix;

transpose(matrix);

matrix.forEach((row) => {

reverse(row, 0, row.length - 1);

});

}

function transpose(matrix) {

for (let i = 0; i < matrix.length; i++) {

for (let j = i; j < matrix[0].length; j++) {

const temp = matrix[i][j];

matrix[i][j] = matrix[j][i];

matrix[j][i] = temp;

}

}

return matrix;

}

function reverse(row, start, end) {

while (start < end) {

const temp = row[start];

row[start] = row[end];

row[end] = temp;

//...

}

//...

}

`

Once the variables are swapped, we want to bring the `start`

and `end`

pointers toward each other, so we'll increment `start`

, and decrement `end`

. Once the while loops is done executing, we can return the now reversed row to `rotate()`

.

``javascript`

function rotate(matrix) {

if (!matrix.length) return null;

if (matrix.length === 1) return matrix;

transpose(matrix);

matrix.forEach((row) => {

reverse(row, 0, row.length - 1);

});

}

function transpose(matrix) {

for (let i = 0; i < matrix.length; i++) {

for (let j = i; j < matrix[0].length; j++) {

const temp = matrix[i][j];

matrix[i][j] = matrix[j][i];

matrix[j][i] = temp;

}

}

return matrix;

}

function reverse(row, start, end) {

while (start < end) {

const temp = row[start];

row[start] = row[end];

row[end] = temp;

start++;

end--;

}

return row;

}

`

Since the problem asked us to rotate the 2D array "in place", we don't have to return anything. We already modified the original matrix, so we're done with our solution!

Let me know in the comments if you have any questions or other ideas for how to approach this problem!

Algorithm of the Day (40 Part Series)

## Discussion