Case 1
Start:
0 1 1
1 1 1
1 1 1

After first scan:
0 1 0
1 1 1
0 1 1

After second scan:
0 1 0
0 1 1
0 1 1

After third scan:
0 0 0
0 1 1
0 1 1

Case 2
Start:
1 1 0
1 1 1
0 1 1

After first scan:
0 1 0
1 1 1
0 1 0

After second scan:
0 0 0
0 1 1
0 0 0

After third scan:
0 0 0
0 1 0
0 0 0

Hope this helps.
Dave

On Nov 22, 9:34 am, "chandra kumar" <[EMAIL PROTECTED]>
wrote:
> Hi Dave,
>     Can you explain your algo for these 2 cases...
>
>     0  1  1            1  1  0
>   1  1  1            1  1  1
>   1  1  1            0  1  1
>
>     Please explain me in steps cause we tried the same problem and can't get
> it done for without using extra space. ( we used 1 extra space, if the
> array can contain only 0 or 1 and no other number, otherwise we store some
> other number indicating a yet to become zero and proceed, in the later case
> we need no extra space )
> Thanks and Regards,
> K.V.Chandra Kumar
>
> On 15/11/2007, Dave <[EMAIL PROTECTED]> wrote:
>
>
>
>
>
> > Scan the array. If the i,j element is zero, set the last element of
> > row i and the last element of column j to zero. Then, scan the last
> > row, ignoring the last element, and zero every column that ends with a
> > zero. Similarly, scan the last column, ignoring the last element, and
> > zero every row that ends with a zero. This touches every element at
> > most 3 times; i.e., if the array has m rows and n columns, the
> > algorithm is O(1) in space and O(m*n) in time.
>
> > Dave
>
> > On Nov 14, 1:27 pm, geekko <[EMAIL PROTECTED]> wrote:
> > > Given an array of 0's and 1's whenever you encounter an 0 make
> > > corresponding column and row elements 0. How could you do that
> > > efficiently(minimum time and minimum space complexity)? This question
> > > is taken from placementsindia.blogspot.com- Hide quoted text -
>
> - Show quoted text -
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