Hi Kumar,
if the array is a bit array, my solution is first copy the entire
array to the auxiliary memory, where -1 can be stored, then do the
other work.
and there's another better solution I memtioned on the above post,
it dosen't require the array to store -1.
Best Regards,
James Fang
On 11月27日, 上午11时35分, "chandra kumar" <[EMAIL PROTECTED]>
wrote:
> Hey,
> You assumed that you can store *-1*, but that is not possible in a bit
> array (i.e., if the question states that the array can contain either 0 or 1
> and nothing more than that). But if the array can contain other numbers as
> you assumed then your solution will work. But the solution that Dave
> mentioned seemed to be assuming that the array cannot contain any other
> element and still need not require any extra memory. That's why I'm posting
> all my doubts to him. If this thread can come up with a solution like that
> then it is really good.
>
> Thanks and Regards,
> K.V.Chandra Kumar.
>
> On 26/11/2007, James Fang <[EMAIL PROTECTED]> wrote:
>
>
>
> > *The below algorithm cost 3*O(m*n)+NumberOfInitialZeros*( O(m)+o(n) )*
>
> > * *
>
> > =================================================================================================================================================================
>
> > No matter the array can hold other elements or not, you can always use
> > auxiliary memory space to make it store more than 0&1, suppose it is a M*N
> > array,
>
> > initializing the auxiliary space will cost O(m*n).
>
> > and the algorithm can be described as follow:
>
> > for(int i=0; i<n; ++i)
>
> > for(int j=0; j<m; ++j)
>
> > {
>
> > if( array[i][j] == 0)
>
> > {
>
> > for(int k=0;k<n;++k)
>
> > {
>
> > if( array[k][j] != 0 ) array[k][j]=-1;
>
> > }
>
> > for(int l=0;l<m;++l)
>
> > {
>
> > if( array[i][l] != 0 ) array[i][l]=-1;
>
> > }
>
> > array[i][j] = -1;
>
> > }
>
> > }
>
> > for(int i=0;i<n;++i)
>
> > for(int j=0;j<m;++j)
>
> > if(array[i][j]==-1)
>
> > array[i][j]=0;
>
> > *Best Regards,** **
> > ** **James Fang*
> > ------------------------------
>
> > *发件人:* algogeeks@googlegroups.com [mailto:[EMAIL PROTECTED] *代表
> > *chandra kumar
> > *发送时间:* 2007年11月23日 21:58
> > *收件人:* algogeeks@googlegroups.com
> > *主题:* [algogeeks] Re: NxN matrix
>
> > Hi, as I understood the question goes like this (I also face this problem
> > earlier)...
>
> > * Given an array of 0s and 1s (whether the array can contain any other
> > element apart from 0 and 1 is not mentioned in the question posted, but to
> > make it hard people say that the array cannot hold element other then 0 and
> > 1, basically a bit array ) . You need to change the given array such that
> > each row and column that have a 0 initially should have all 0s in it.*
>
> > * e.g.*
>
> > * 1 1 1 1 1 **0** 1 1*
>
> > * 1 0 1 1 **0 **0** 0 0*
>
> > * 1 1 1 1 ====> 1 **0** 1 1*
>
> > * 1 1 1 1 1 **0** 1 1*
>
> > Thanks and Regards,
>
> > K.V.Chandra Kumar
>
> > On 23/11/2007, *MJ* <[EMAIL PROTECTED]> wrote:
>
> > HI Chandra,
>
> > As per the problem statement
>
> > "Given an array of 0's and 1's whenever you encounter an 0 make
> > corresponding column and row elements 0. "
>
> > Does it mean make all the rows and column zero or only last row and
> > column elemnt zero??
> > could you reframe the problem statement with more details or an
> > example.
>
> > Thank YOu,
> > Mayur
>
> > On Nov 23, 1:20 pm, "chandra kumar" <[EMAIL PROTECTED]>
> > wrote:
> > > I have some questions in case 2.... questions inlined..
>
> > > Thanks and regards,
> > > K.V.Chandra kumar.
>
> > > On 22/11/2007, Dave <[EMAIL PROTECTED]> wrote:
>
> > > > 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 how did we get 0 in the first row first column, we
> > > > should only alter the last row and column, right?
> > > > 1 1 1
> > > > 0 1 0
>
> > > > After second scan:
> > > > 0 0 0 how did we get these zero, only columns are to be
> > > > changed, right?
> > > > 0 1 1
> > > > 0 0 0
>
> > > > After third scan:
> > > > 0 0 0
> > > > 0 1 0 how did we get this as, ast element is ignored,
> > right?
> > > > 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 -- Hide quoted text -
>
> > > - Show quoted text -
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