Here is a pseudocode for one of the solutions sumN = sum of all the elements of the arraysumN2 = sum of 1 to (n+2) sumofXY = sumN2 - sumN /*sum of the interested integers, say x, y*/ xorN = XOR of all the elements of the array xorN2 = XOR of 1 to (n+2) xorofXY = xorN XOR xorN2 /*XOR of x, y*/ for i=1 to sumofXY/2 if (xorofXY XOR i XOR (sumofXY-i)) is zero then x is i, y is (sumofXY-i) endfor
This solution has running time complexity of O(n) and needs constant extra space. Thanks, Channa On Wed, Jul 29, 2009 at 5:57 PM, Ajinkya Kale <[email protected]> wrote: > use hashing. > > On Wed, Jul 29, 2009 at 4:50 PM, Vijayasarathy K > <[email protected]>wrote: > >> >> Consider an array of 'n' elements which contains all except 2 numbers >> from 1....(n + 2). How can we find the 2 missing elements? >> >> >> > > > -- > Ciao, > Ajinkya > > > > > --~--~---------~--~----~------------~-------~--~----~ 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 -~----------~----~----~----~------~----~------~--~---
