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Sockets - Cyndi #6
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Original file line number | Diff line number | Diff line change |
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@@ -3,8 +3,35 @@ | |
# If any number is found to be 0, the method updates all the numbers in the | ||
# corresponding row as well as the corresponding column to be 0. | ||
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||
# Time complexity: ? | ||
# Space complexity: ? | ||
# Time complexity: the time complexity is O(n*m) where n is the number of rows in the matrix and m is the number of columns in the matrix because I have two nested loops O(2*n*m), which dropping the constant reduces to O(n*m) | ||
# Space complexity: space complexity is O(n) where n is the larger number of rows or columns. This is because a hash is created where the rows are keys and columns are values | ||
def matrix_convert_to_zero(matrix) | ||
raise NotImplementedError | ||
num = 0 | ||
pos = Hash.new | ||
rows = matrix.size | ||
columns = matrix[0].size | ||
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||
rows.times do |row| | ||
columns.times do |col| | ||
if matrix[row][col] == 0 | ||
if pos.include? row | ||
num += 1 | ||
pos["row" + num.to_s] = col | ||
else | ||
pos[row] = col | ||
end | ||
end | ||
end | ||
end | ||
rows.times do |row| | ||
columns.times do |col| | ||
if pos.include? row | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. I suggest reducing the two if statements to: However remember the .include? is a loop, so this is another nested loop. |
||
matrix[row][col] = 0 | ||
end | ||
if pos.values.include? col | ||
matrix[row][col] = 0 | ||
end | ||
end | ||
end | ||
return matrix | ||
end |
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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.