> > > I am sceptical whether any XOR solution exits for your question. But if > the question is modified as : > > *Only one number repeats once,* some no.s repeat twice and only one number > repeat thrice, here is the XOR solution for that. > > suppose the sample array is A[]={1, 3,3,5,5,5, 7,7,8,8} > in the example 1 repeats once and 5 repeats thrice. > > 1>let T= XOR( all elements)= 1^5. (all elements occurring even no of times > nullify) -O(N) > > ( let x=1, y=5 > As we know the no. repeating once and the no. repeating thrice are unequal, > there must exist some bit 'k' such that x[k]!=y[k]. There may be more than > such bits in x and y. But one such bit certainly yields T[k]=1 after x^y) > > 2> Now traverse along each bit of T( in binary) from left or right and > consider T[i] =1 which is encountered first. store it . let b=i; > (O(M) time and O(M) space to store binary if M is the bit length of T.) > > 3> T1= XOR(all elements in given array having bit b as 1)..... (O(N) time > and O(M) space) ( time is O(MN) but as M<=32 , complexity remain O(N)) > > 4> T0= XOR( all elements in given array having bit b as 0) (O(N) time and > O(M) space) > > One of (T1,T0) gives the no. that repeats once and the other gives the no > that repeats thrice. > > 6> Now traverse the along array A and compute count for T1 and T0. The > count that equals 3 gives the corresponding no. repeating thrice. -O(N) > > Time complexity is O(N+M) . Linear > space complexity is O(M) to store binary form. > > But this algo certainly fails if more than one no. repeats once. > > > -- Thanks & Regards, Priyanka Chatterjee Final Year Undergraduate Student, Computer Science & Engineering, National Institute Of Technology,Durgapur India http://priyanka-nit.blogspot.com/
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