@pacific in your case a[0][0] might be wrong. So lets do this :
a[0][0] = a[0][0 to n-1] & a[1 to n-1][0] // & of all elements in first row
and first column
for( int i=1; i<n; i++)
{
for( int j=1;j<n;j++)
{
a[0][i] &= a[j][i]; // a[0][i] contains & of all elements
of ith column
a[i][0] &= a[i][j]; // a[i][0] contains & of all elements
of ith row
}
}
for( int i=1; i<n; i++)
{
for( int j=1;j<n;j++)
{
a[i][j] = a[i][0] & a[0][j] ;
}
}
On Sun, Feb 27, 2011 at 6:37 PM, gaurav gupta <[email protected]>wrote:
> @pacific Absolutely correct.
>
> Thanks for the explanation.
>
>
> 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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>
>
>
> --
> Thanks & Regards,
> Gaurav Gupta
>
> "Quality is never an accident. It is always result of intelligent effort" -
> John Ruskin
>
>
--
Thanks & Regards,
Gaurav Gupta
7676-999-350
"Quality is never an accident. It is always result of intelligent effort" -
John Ruskin
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