Can use a map to read the array and do map[arr[i]]=1. If there's a 1 already present print arr[i]. This method solves the problem in O(n) time and O(n) space complexity.
On Jul 29, 2:56 am, Minotauraus <anike...@gmail.com> wrote: > If numbers are consecutive you can do better. Calculate (n-1)(n-2)/2 - > sum of all elements in the array. > > That'll require one less loop compared to XORing twice. > > -Minotauraus. > > On Jul 28, 8:51 am, Apoorve Mohan <apoorvemo...@gmail.com> wrote: > > > Solution : > > > 1. Find Xor of numbers from 1 to n-1. > > > 2. Find Xor of the numbers present in the array. > > > 3. Xor the results from step 1 and 2 you will get the repeated number. > > > On Wed, Jul 28, 2010 at 8:46 PM, akshay <akshayrastogi2...@gmail.com> wrote: > > > An array of unsorted numbers n is given with one no.repeated once ie > > > n-1 distinct nos to find duplicate no. in o(n) complexity > > > > -- > > > You received this message because you are subscribed to the Google Groups > > > "Algorithm Geeks" group. > > > To post to this group, send email to algoge...@googlegroups.com. > > > To unsubscribe from this group, send email to > > > algogeeks+unsubscr...@googlegroups.com<algogeeks%2bunsubscr...@googlegroups.com> > > > . > > > For more options, visit this group at > > >http://groups.google.com/group/algogeeks?hl=en. > > > -- > > regards > > > Apoorve Mohan -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
