48. rotate image

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.

solutions

logical

Literally just matrix logic. We keep four pointers starting on each corner of each layer, and move our pointers along the sides.

def rotate(self, matrix: List[List[int]]) -> None:
	"""
	Do not return anything, modify matrix in-place instead.
	"""
	n = len(matrix)
	its = n // 2
	
	for it in range(its):
		tl = [0 + it, 0 + it]
		tr = [0 + it, n - 1 - it]
		bl = [n - 1 - it, 0 + it]
		br = [n - 1 - it, n - 1 - it]
		
		for _ in range(n-1-(2*it)):
			tr_temp = matrix[tr[0]][tr[1]]
			matrix[tr[0]][tr[1]] = matrix[tl[0]][tl[1]]
			matrix[tl[0]][tl[1]] = matrix[bl[0]][bl[1]]
			matrix[bl[0]][bl[1]] = matrix[br[0]][br[1]]
			matrix[br[0]][br[1]] = tr_temp
			
			tl[1] += 1
			tr[0] += 1
			br[1] -= 1 
			bl[0] -= 1

reverse + reflect

Kinda just have to draw this one out…

def rotate(self, matrix: List[List[int]]) -> None:
	"""
	1. reverse
	2. reflect
	for an index (i, j), reflection index is (n-1-j, n-1-i)
	"""
	  
	n = len(matrix)
 
	def reverse(row):
		l, r = 0, n-1
		while l < r:
			row[l], row[r] = row[r], row[l]
			l += 1
			r -= 1
	  
	# reverse
	for row in matrix:
		reverse(row)
 
	# reflect over /
	for i in range(n-1):
		for j in range(n-i-1):
			matrix[i][j], matrix[n-j-1][n-i-1] = matrix[n-j-1][n-i-1], matrix[i][j]