Hi MD,
In your last approach, watch out for arbitrarily large values. What
would 'N' be? By the way, 'for' loop there should be for 'n', which is the
size of array 'a'. A simple array like {20, 5, 123456789} would kill your
memory.
Hashing approach would be great if we've some idea on nature of the array
contents. Here's a method of O(n logn) again.
Parse through each entry in the array and insert into a BST (Binary Search
Tree). Ignore duplicate entries and hence gives unique elements when done
in-traversal later. However, worst-case shoots to O(n^2) if the distribution
is skewed.
Regards,
Channa
On Oct 30, 2007 5:01 AM, MD <[EMAIL PROTECTED]> wrote:
>
> There can be multiple ways to do this.. obviously sorting gives you
> o(nlogn).
> another approach of hashtable gives o(n)
> and here is another approach:
> 1. maintaining array of size N, initialized to 0. Say B
> 2. for(i=0;i<N;i++) { B[a[i]]++;}
> 3. Output all values with B[i] >= 1 , which gives unique elements
> only.
>
> Only restriction is the memory.
>
>
> >
>
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