Algorithm:
1. Given array A[], Set B[], sum=0,repeatedNumb=0;
2. for index i to n
B[ A[i] -1 ] = B[ A[i] -1]+1
sum=sum+A[i]
3. for index i to n
if B[i]=2
repeatedNumb=i+1
break;
4. Nsum=n*(n+1)/2
5. missingNumb=Nsum+repeatedNumb-sum
Step-2 to3, finding the repeatedNum, complexity=O(n)+O(n)
Step4: finding the Nsum, complexity=O(n) if formula is not used
Thanks,
Krishna
On Jul 18, 6:26 pm, SkRiPt KiDdIe <anuragmsi...@gmail.com> wrote:
> > @Nishant:
>
> >>> 1 4 5 4 5 n=5
>
> >>> 1 2 3 4 5
>
> >>> after xor i.e. your x1 answer contains (2^3^4^5).The missing elements are
> >>> included in xor as well along with repeating elements.
>
> >>> Hope now you got it. You are giving solution for a question which i have
> >>> defined in previous post. and your algo will fail when
> >>> the final xor has no set bit i.e. same number is being repeated twice.
>
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