Question

Given a NxN matrix with 0s and 1s. Set every row that contains a 0 to all 0s and set every column that contains a 0 to all 0s.

For example

1 0 1 1 0 0 1 1 1 0 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 

results in

0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 1 1 0 

A Microsoft Engineer told me that there is a solution that involves no extra memory, just two boolean variables and one pass, so I’m looking for that answer.

BTW, imagine it is a bit matrix, therefore just 1s and 0s are allow to be in the matrix.

Answer 1

Ok, so I’m tired as it’s 3AM here, but I have a first try inplace with exactly 2 passes on each number in the matrix, so in O(NxN) and it is linear in the size of the matrix.

I use 1rst column and first row as markers to know where are rows/cols with only 1’s. Then, there are 2 variables l and c to remember if 1rst row/column are all 1’s also. So the first pass sets the markers and resets the rest to 0’s.

The second pass sets 1 in places where rows and cols where marked to be 1, and resets 1st line/col depending on l and c.

I doubt strongly that I can be done in 1 pass as squares in the beginning depend on squares in the end. Maybe my 2nd pass can be made more efficient…

import pprint  m = [[1, 0, 1, 1, 0],      [0, 1, 1, 1, 0],      [1, 1, 1, 1, 1],      [1, 0, 1, 1, 1],      [1, 1, 1, 1, 1]]    N = len(m)  ### pass 1  # 1 rst line/column c = 1 for i in range(N):     c &= m[i][0]  l = 1 for i in range(1,N):     l &= m[0][i]   # other line/cols # use line1, col1 to keep only those with 1 for i in range(1,N):     for j in range(1,N):         if m[i][j] == 0:             m[0][j] = 0             m[i][0] = 0         else:             m[i][j] = 0  ### pass 2  # if line1 and col1 are ones: it is 1 for i in range(1,N):     for j in range(1,N):         if m[i][0] & m[0][j]:             m[i][j] = 1  # 1rst row and col: reset if 0 if l == 0:     for i in range(N):         m [i][0] = 0  if c == 0:     for j in range(1,N):         m [0][j] = 0   pprint.pprint(m)