Set every cell in matrix to 0 if that row or column contains a 0

jaircazarin-old-account picture jaircazarin-old-account · Dec 4, 2008 · Viewed 53.5k times · Source

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

Piotr Lesnicki picture Piotr Lesnicki · Dec 4, 2008

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)