@Vivek
I had told abt tat border case already once..
Suppose the two missin numbers are greater than n, then m==0 when exitin the
loop.
So they will be n+1 and n+2 only.
in case, one of the missin numbers is greater than n, then m==1, and can be
simply found by subtracting the (array_sum+x[0] )
@all
suming the 1 to n+2 or n will not solve this problem..
it may give overflow problem..
this is not the soln...
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we have got additional data typees to handle sir,like biginteger in
java.string in c++ etc.its a linear programming problem sir.
On Sat, Aug 1, 2009 at 6:12 PM, ankur aggarwal ankur.mast@gmail.comwrote:
@all
suming the 1 to n+2 or n will not solve this problem..
it may give overflow
@sharad
then wat about memory wastage??
memory shud be taken care of..
On Sat, Aug 1, 2009 at 7:39 PM, sharad kumar aryansmit3...@gmail.comwrote:
we have got additional data typees to handle sir,like biginteger in
java.string in c++ etc.its a linear programming problem sir.
On Sat, Aug
yes sir,bigint can handle till 100! or more .incase of overflow we can make
use of string
On Sat, Aug 1, 2009 at 8:00 PM, ankur aggarwal ankur.mast@gmail.comwrote:
@sharad
then wat about memory wastage??
memory shud be taken care of..
On Sat, Aug 1, 2009 at 7:39 PM, sharad kumar
Try {2, 1}
On Sat, Aug 1, 2009 at 11:45 AM, Devi G devs...@gmail.com wrote:
@Vivek
I had told abt tat border case already once..
Suppose the two missin numbers are greater than n, then m==0 when exitin
the loop.
So they will be n+1 and n+2 only.
in case, one of the missin numbers is
