Problem: Detecting duplicated items in an array

And another problem :-).

Someone asked me to give him problems so that he can train for an interview. I gave him the problems from google posted here and in turn he gave me another one: You have an array of length n. It is filled with n random numbers between 0 and n-1. The task it to find out if any of them are repeated.

The trivial solution would be to sort the array and detect any repeated number in the sorted array. This one has O(n * log(n)) time complexity. We should be able to do better. Here is what i came out with after a little thinking.

• start with the first element of the array which has a value different from the index at which it is located. For int[] arr = {0, 1, 2, 3, 5, 4} i would start with the position 4 because the value of 5 is different than its index.
• take the value and go in the array at the position represented by that index. Look at the value from there. If the value is the same as that index we just found our duplicate. If not then remember the new value, put the old value in there and do the same.
• if the next value is the original starting position you have just arranged a subset of the original array in the proper place. select the next number matching the condition at step 1 if possible. If not then you are done, you have the array sorted and you also know that there are no repeted numbers in it.

The worst case complexity is lower than O(2 * n). In the best case and medium case probably you will find in less than n/2 operations but this is just a guess. You would have to apply some statistics :-) to find out exactly what is this in the medium case.

Mihai.

Goooogle interview questions

The other day I got hooked by a friend of mine with a couple of interview questions given by google. Some of them are challenging and quite interesting. Here is one (taken from here): Question #8) Given an array A[string], an array of strings where each string represents a word in a text document. Also given 3 search terms T1, T2, and T3 and 3 corresponding sorted sequences of integers S1, S2, and S3 where each integer in Si represents an index in A where search term Ti occured (i.e. S1, S2, and S3 contain the locations of the search terms in our array of words). Now find a minimal subarray of A that contains all of the search terms T1, T2, and T3. Extend this algorithm for an arbitrary number of search terms. This one in effect is asking you to solve the excerpting problem: Given a text, a couple of search terms with the matching positions in the original text you are asked to return an "interesting" fragment of the original text containing all the search term. In this case the interesting criteria is the length of the fragment: the fragment is most interesting when the length is shorter.

My first solution was to the [max(min(S1), min(S2), min(S3)), min(max(S1), max(S2), max(S3))] which is not good. The reason is that you can have cases in which the interval you choose can have a one or more search terms left out: S1 = {1, 3}, S2 = {5, 7}, S3 = {4, 8}. The interval you would get in this case would have been 1, 3 which will only contain the first term. You can also choose the min(min()..) .. max(max()..) solution which is good as an interval containing all the terms but is too broad (in effect is the longest possible). You can merge the S1, S2, S3 sequences and sort them (having "colored" the indexes first). This will be a proper sequences (it has all the colors in it) but it is too long. We need to shorten it. For this we will look at the sequence ends for the shortest subsequences beginning and ending at each end of the current sequence. For example given this sequence: [R1, B3, Y4, R5, .... R11, B10, B15, Y16] we will look at those 2 subsequences: [R1, B3, Y4], [R11, B10, B15, Y16]. Right now we have 3 sequences: the original, the first one of length 3 and the second one of length 4. Because we are looking for the sequence of minimal length we can break the one with length 4. We will ignore then the Y16 term and go ahead and rebuild the second sequence looking deeper into the original sequence. Once we find a proper one we apply the same rule (break the longest one). This will guarantee that we will always have the shorter one. The algorithm will stop when both the found sequences are in fact the same.

There is a problem somewhere in this algorithm but I'll let a little fun in problem for the readers :-). You can easily extend this to more than 3 terms maintaining the exact complexity. The actual complexity of this is O(n) where n is the sum of the term positions in the original text. Also there is another solution totally different than this one and quite interesting which for multiple terms might be faster.
Maybe I'll post it at some point.

Mihai