@Sachin: You can sort the array in O(n log n), and then finding the
two numbers is an additional O(n), so the resulting complexity is O(n
log n).
Dave
On Aug 25, 2:01 pm, sachin sabbarwal <algowithsac...@gmail.com> wrote:
> We have an array which contains integer numbers (not any fixed range). Only
> two numbers are repeated odd number of times and remaining even number of
> times. Find the 2 numbers.
>
> like
>
> arr[]={1,2,5,1,5,1,1,3,2,2}
>
> 1 -> 4 times(even)
> 2-> 3 times(odd)
> 3-> 1 times(odd)
> 5-> 2 times(even)
>
> output 2 3
>
> m not able to come up with a non-hashing solution which is faster than n^2,
> plz suggest both type of solutions viz. hashing and non-hashing one's.
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