Sockets - Cyndi #6
Conversation
There was a problem hiding this comment.
This definitely works, but because you're using .includes your time complexity is actually O(n^3). Instead I suggest using 2 hashes so you can just look up and check to see if the index is in the hash with O(1).
There is also a counterintuitive way to solve the problem without the sets you used, instead using the 1st row and 1st column to store the row and columns to zero out. However this is a good solution. Nice work.
| end | ||
| rows.times do |row| | ||
| columns.times do |col| | ||
| if pos.include? row |
There was a problem hiding this comment.
I suggest reducing the two if statements to: if pos.include? row || pos.values.include? col
However remember the .include? is a loop, so this is another nested loop.
| rows.times do |row| | ||
| columns.times do |col| | ||
| if matrix[row][col] == 0 | ||
| if pos.include? row |
There was a problem hiding this comment.
I'm not sure the point of this if statement.
I suggest instead of this if-else statement just having two hashes, one for the rows and the other for columns.
No description provided.