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Given a square matrix mat[][], Rotate the matrix by 180 degrees.
Note: Rotating 180° clockwise or anticlockwise gives the same result.
Examples:
Input: mat[][] = [[1, 2, 3],
[4, 5, 6],
[7, 8, 9]]Output: [[9, 8, 7],
[6, 5, 4],
[3, 2, 1]]Explanation: The output matrix is the input matrix rotated by 180 degrees.
Input: mat[][] = [[1, 2, 3, 4],
[5, 6, 7, 8],
[9, 0, 1, 2],
[3, 4, 5, 6]]Output: [[6, 5, 4, 3],
[2, 1, 0, 9],
[8, 7, 6, 5],
[4, 3, 2, 1]]Explanation: The output matrix is the input matrix rotated by 180 degrees.
Table of Content
A simple solution is to use the solutions discussed in Rotate 90 Degree Counterclockwise or Rotate 90 Degree Clockwise two times. These solutions require more effort. We can solve this problem more efficiently by directly finding a relationship between the original matrix and 180 degree rotated matrix.
If we take a closer look at the examples, we can notice that after rotation, the first element in the top row moves to the last cell in the last row. Similarly, second element in the top row moves to the second last cell in the last row, and so on. In general, we can notice that mat[i][j] needs to be placed at cell [n-i-1][n-j-1]. So, we can create a new matrix and place all the elements at their correct position. Finally, we copy all the elements from new matrix to the original matrix.
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In the above approach, we are placing mat[i][j] at cell [n-i-1][n-j-1]. However, instead of placing mat[i][j] at cell [n-i-1][n-j-1] in a new matrix, we can observe that mat[n-i-1][n-j-1] is also being placed back at mat[i][j]. So, rather than performing this in two separate matrices, we can handle it in same matrix by simply swapping mat[i][j] and mat[n-i-1][n-j-1].
If the matrix has an odd number of rows, the middle row won’t have an opposite to swap with, so we need to handle it separately. This middle row elements have to be reversed among themselves.
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