hi this problem is already solver previously. check for "Re: Finding Repeated and Missing Numbers"
if X is the missing number and Y is the repeating number. then let S(N) = n*(n+1)/2 sum of all the numbers provided be Sn ler S(N2) = n*(n+1)*(2*n+1)/6 sum of square of all the numbers provided be Sn2 now if D1 = S(N) - Sn and D2 = S(N2) - Sn2 then X = (D2 + (D1*D1)) / (2*D1) and Y = (D2 - (D1*D1)) / (2*D1) So its a order O(n) solution. regards swadhin --~--~---------~--~----~------------~-------~--~----~ 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 -~----------~----~----~----~------~----~------~--~---
