Finding all unique elements from a big sorted array in log n time?

Question

I have a very big sorted array. How can I count or print all the unique elements of an array??

Suppose my array is [2,3,3,3,4,6,6,7] then output should be 2,3,4,6,7

I know to do it in a n(complexity) time. But interviewer asked me to do this in log n time?? Is it possible?

Since printing elements is O(n) complexity it shouldn't be unless you have only a predefined number of different elements (suppose you have integer numbers 1 to 10 in the array). — George Nechifor
– George Nechifor, Commented Jun 19, 2014 at 8:32
Interviewer is not asking to do task with (log n) number of operations. He is asking the complexity to be (log n) it means number of operations are proportional to (log n). e.g. for o(n) algo if for 100 elements it takes 100,000 operations then for 200 elements it will take 200,000 operations. where as for o(log n) algo if for 100 elements it takes 100,000 operations then for 200 elements it will take 100,150 etc. (Ignore actual numbers. They are just indicative) — Kaushik Lele
– Kaushik Lele, Commented Jun 21, 2014 at 11:13

Vikram Bhat · Accepted Answer · 2014-06-19 09:02:45Z

7

Here is an algorithm which requires O(logn*k) where k is unique elements:-

set uniQ
int ind = 0;

do {

 uniQ.add(arr[i]);
 ind = BinSearchGreater(arr,arr[ind],ind+1);
 if(ind >= arr.length)
   break;

} while(true);


BinSearchGreater(arr,key,start_ind) : returns index of first element greater than key in subarray starting at start_ind

Time complexity :- Note this algorithm is only good when no of unique elements are small. This is asymptotically O(n*logn) if all are unique so worse than linear.

answered Jun 19, 2014 at 9:02

Vikram Bhat

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1 Comment

Niklas B. Over a year ago

My interviewer was very pleased with this solution when I was asked the same question once.

Flupp · Accepted Answer · 2014-06-19 08:31:15Z

0

I would like to know how he (the interviewer) counts every unique element in the array [1,2,3,4,5] without picking at least every element. In this case you have to pick every element to count every element and this will be done in O(n). In my opinion impossible to get a complexity of O(log n), if there are no other requirements to the given array.

answered Jun 19, 2014 at 8:31

Flupp

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9 Comments

perreal Over a year ago

the array is sorted so you don't really need to look at every element.

piotrekg2 Over a year ago

You have to at least read that whole array so lower bound is O(n)

Pham Trung Over a year ago

@perreal you have to look at every element, for example, this case (2,5,7,100) how can you tell?

piotrekg2 Over a year ago

I think it can be done in M log N where M is size of the result. You can achieve this by running multiple binary searches.

perreal Over a year ago

@PhamTrung, that is a list of all uniq elements so even printing is O(n). What if you have log(n) unique elements?

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