Alkesh

LeetCode - Rotate Image

Container

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.

Container

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;
        }
    }
};

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]]
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