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

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]

l = 1
for i in range(1,N):
l &= m[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[j] = 0
m[i] = 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] & m[j]:
m[i][j] = 1

# 1rst row and col: reset if 0
if l == 0:
for i in range(N):
m [i] = 0

if c == 0:
for j in range(1,N):
m [j] = 0

pprint.pprint(m)
``````