# LeetCode - Rotate Image

## Problem statement

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.

Problem statement taken from: https://leetcode.com/problems/rotate-image

Example 1:

``````Input: matrix = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
Output: [[7, 4, 1], [8, 5, 2], [9, 6, 3]]
``````

Example 2:

``````Input: matrix = [[5, 1, 9, 11], [2, 4, 8, 10], [13, 3, 6, 7], [15, 14, 12, 16]]
Output: [[15, 13, 2, 5], [14, 3, 4, 1], [12, 6, 8, 9], [16, 7, 10, 11]]
``````

Example 3:

``````Input: matrix = [[1]]
Output: [[1]]
``````

Example 4:

``````Input: matrix = [[1, 2], [3, 4]]
Output: [[3, 1], [4, 2]]
``````

Constraints:

``````- matrix.length == n
- matrix[i].length == n
- 1 <= n <= 20
- -1000 <= matrix[i][j] <= 1000
``````

### Explanation

#### Rotate Groups of Four Cells

The first solution we can think of is to rotate the four corners of the matrix. We repeat this of the subsequent cells too.

Let's check the algorithm.

``````- initialize m = matrix.size, tmp

- loop for i = 0; i < m / 2; i++
- loop for j = i; j < m - 1 - i; j++
- tmp = matrix[i][j]
- matrix[i][j] = matrix[m - 1 - j][i]
- matrix[m - 1 - j][i] = matrix[m - 1 - i][m - 1 - j]
- matrix[m - 1 - i][m - 1 - j] = matrix[j][m - 1 - i]
- matrix[j][m - 1 - i] = tmp
``````

The time complexity of the program is O(M) as each cell is getting read once and written once. Space complexity is O(1) because we do not use any other additional data structures.

#### C++ solution

``````class Solution {
public:
void rotate(vector<vector<int>>& matrix) {
int m = matrix.size(), tmp;
int j = 0;

for(int i = 0; i < m/2; i++){
for(int j = i; j < m - 1 - i; j++){
tmp = matrix[i][j];
matrix[i][j] = matrix[m - 1 - j][i];
matrix[m - 1 - j][i] = matrix[m - 1 - i][m - 1 - j];
matrix[m - 1 - i][m - 1 - j] = matrix[j][m - 1 - i];
matrix[j][m - 1 - i] = tmp;
}
}
}
};``````

#### Golang solution

``````func rotate(matrix [][]int)  {
m := len(matrix)
tmp := 0

for i := 0; i < m / 2; i++ {
for j := i; j < m - 1 - i; j++ {
tmp = matrix[i][j];
matrix[i][j] = matrix[m - 1 - j][i];
matrix[m - 1 - j][i] = matrix[m - 1 - i][m - 1 - j];
matrix[m - 1 - i][m - 1 - j] = matrix[j][m - 1 - i];
matrix[j][m - 1 - i] = tmp;
}
}
}``````

#### Javascript solution

``````var rotate = function(matrix) {
let m = matrix.length;
let tmp = 0;
let i, j;

for(i = 0; i < m/2; i++){
for(j = i; j < m - 1 - i; j++){
tmp = matrix[i][j];
matrix[i][j] = matrix[m - 1 - j][i];
matrix[m - 1 - j][i] = matrix[m - 1 - i][m - 1 - j];
matrix[m - 1 - i][m - 1 - j] = matrix[j][m - 1 - i];
matrix[j][m - 1 - i] = tmp;
}
}
};``````

#### Dry Run

Let's dry-run our algorithm to see how the solution works.

``````Input:
matrix = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]

Step 1: m = matrix.length
m = 3

Step 2: tmp = 0

Step 3: loop i = 0; i < 3/2
0 < 3/2
0 < 1
true

j = i; j < 3 - 1 - 0
0 < 2
true

tmp = matrix[i][j]
tmp = 1

matrix[i][j] = matrix[m - 1 - j][i]
matrix[0][0] = matrix[3 - 1 - 0][0]
matrix[0][0] = matrix[2][0]
matrix[0][0] = 7

matrix[m - 1 - j][i] = matrix[m - 1 - i][m - 1 - j]
matrix[3 - 1 - 0][0] = matrix[3 - 1 - 0][3 - 1 - 0]
matrix[2][0] = matrix[2][2]
matrix[2][0] = 9

matrix[m - 1 - i][m - 1 - j] = matrix[j][m - 1 - i]
matrix[3 - 1 - 0][3 - 1 - 0] = matrix[0][3 - 1 - 0]
matrix[2][2] = matrix[0][2]
matrix[2][2] = 3

matrix[j][m - 1 - i] = tmp
matrix[0][3 - 1 - 0] = 1
matrix[0][2] = 1

j++
j = 1

Step 4: j < 2
1 < 2
true

tmp = matrix[0][1]
tmp = 2

matrix[i][1] = matrix[m - 1 - j][i]
matrix[0][1] = matrix[3 - 1 - 1][0]
matrix[0][1] = matrix[1][0]
matrix[0][1] = 4

matrix[m - 1 - j][i] = matrix[m - 1 - i][m - 1 - 1]
matrix[3 - 1 - 1][0] = matrix[3 - 1 - 0][3 - 1 - 1]
matrix[1][0] = matrix[2][1]
matrix[1][0] = 8

matrix[m - 1 - i][m - 1 - j] = matrix[j][m - 1 - i]
matrix[3 - 1 - 0][3 - 1 - 1] = matrix[1][3 - 1 - 0]
matrix[2][1] = matrix[1][2]
matrix[2][1] = 6

matrix[1][m - 1 - i] = tmp
matrix[1][3 - 1 - 0] = 1
matrix[1][2] = 2

j++
j = 2

Step 5: j < 2
2 < 2
false

Step 6: i++
i = 1

1 < 3/2
1 < 1
false

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