The above solution will work if each other number is exactly once present. but if that 's not true. then 4 equations can be formed. Assuming a,b repeated number where a may or may be equal to b.
then equations will be x + y = a + b; x^2 + y^2 = a^2 + b^2. x.y = a.b x^3 + y^3 = a^3 + b^3. now 4 equations 4 variables can be solved. On Mon, Jul 18, 2011 at 5:31 PM, ankit sambyal <[email protected]>wrote: > 1. Initialize a bit vector of size n. > 2. For every no. set the corresponding bit vector. > 3. Now scan through the bit vectors and get the missing numbers > corressponding to the unset bits in the bit vector. > > > Time complexity : O(n) > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to [email protected]. > To unsubscribe from this group, send email to > [email protected]. > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > > -- Varun Pahwa B.Tech (IT) 7th Sem. Indian Institute of Information Technology Allahabad. Ph : 09793899112 Official Email :: [email protected] Another Email :: [email protected] People who fail to plan are those who plan to fail. -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
