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)
```