CLEAN CODE: SPIRAL ORDER MATRIX
By: Saurav
2018-01-22 10:49:00 UTC
Given an integer n, generate a square matrix filled with elements from 1 to n2 in spiral order.
Example:
Given n = 3,
You should return the following matrix:
[
[ 1, 2, 3 ],
[ 8, 9, 4 ],
[ 7, 6, 5 ]
]
This problem is a pattern based problem. Its more of a how you use while and for loop rather than complex logic or algorithm.
Algo:
The spiral order filling of array elements can be done by keeping four buffer values which keep track of the start x, end x, start y and end y while filling up the
buffer array with values.
When you fill up a single dimension array inside the two square array, you need to update the variables to the next row or column.
It would be more clear walking through the code as below:
let's start with writing some tests:
p generateMatrix(0) == [] p generateMatrix(1) == [1] p generateMatrix(2) == [[1, 2], [4, 3]] p generateMatrix(3) == [[1, 2, 3], [8, 9, 4], [7, 6, 5]] p generateMatrix(4) == [[1, 2, 3, 4], [12, 13, 14, 5], [11, 16, 15, 6], [10, 9, 8, 7]] p generateMatrix(5) == [[1, 2, 3, 4, 5], [16, 17, 18, 19, 6], [15, 24, 25, 20, 7], [14, 23, 22, 21, 8], [13, 12, 11, 10, 9]]
def generateMatrix(a)
return Array.new if a < 1
return Array.new(1,1) if a < 2
buffer_array = Array.new(a){Array.new(a, 0)}
start_x = 0
start_y = 0
end_x = a - 1
end_y = a - 1
i = 1
max = a**2
while i <= max
(start_x..end_x).each do |k|
buffer_array[start_y][k] = i
i += 1
end
start_y += 1
(start_y..end_y).each do |k|
buffer_array[k][end_x] = i
i += 1
end
end_x -= 1
(end_x).downto(start_x).each do |k|
buffer_array[end_y][k] = i
i += 1
end
end_y -= 1
(end_y).downto(start_y).each do |k|
buffer_array[k][start_x] = i
i += 1
end
start_x += 1
end
return buffer_array
end
Works fine!
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