If the diagonal elements of the matrix are all 0s, then you'd have to set
every element in the matrix to 0 (i.e. O(N^2) operations ). I don't think,
therefore, that we can do better than O(N^2). The best we can do is to
perhaps, make it output sensitive.




On Sun, Feb 27, 2011 at 6:14 PM, pacific pacific <[email protected]>wrote:

> 1. do & operation on all the values in each column and store it in the
> first row of each column
> 2. do & operation on all the values in each row and store it in the first
> column of each row.
> (when writing at a[0][0] do & operation with the value computed at 1.)
> 3. Now to find out the value at a[i][j] ,you need to do a[i][0] & a[0][j]
>
> On Sun, Feb 27, 2011 at 12:03 PM, Rajnish <[email protected]> wrote:
>
>> 1.) Traverse the whole matrix and replace each 0 value with -1.
>> 2.) Traverse the matrix again,all the 1 values are replaced with 0 in
>> the row and column of the index where a -1 value is found.
>> 3.) Set all -1 values to zero and we have the output array.
>> time complexity: O(n^2)
>> space complexity: O(1)
>>
>>
>> On Feb 27, 2:29 am, gaurav gupta <[email protected]> wrote:
>> > A NxN binary matrix is given. If a row contains a 0 all element in the
>> > row will be set to 0 and if a column contains a 0 all element of the
>> > column will be set to 0. You have to do it in O(1) space.
>> >
>> > example :
>> >
>> > input array :
>> >
>> > 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
>> >
>> > result array :
>> >
>> > 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
>> >
>> > Thanks & Regards,
>> > Gaurav Gupta
>>
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>
>
> --
> regards,
> chinna.
>
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