# Google Interview Question: Product of other Elements in an Array in O(n)

Tags: , | Posted by Admin on 6/10/2008 1:25 PM | Comments (0)

Last time I was interviewed for a software development engineer position, the recruiter asked me some of the classical Microsoft interview questions, such as “How Would You Move Mount Fuji?” or “How many gas station are there in your country?“.

It was the first time for me to be asked such questions but having obtained the job I think my answers were good enough.

After that day, I looked for other well-known interview questions and I discovered that Google has a totally different approach, focusing on algorithms, data structures and complexity.

For instance, one of Google interview questions says:

There is an array A[N] of N integers. You have to compose an array Output[N+1] such that Output[i] will be equal to the product of all the elements of A[] except A[i].

Example:
INPUT:[4, 3, 2, 1, 2]
OUTPUT:[12, 16, 24, 48, 24]

Solve it without division operator and in O(n).

Without the two constraints at the end of the puzzle it would have been straightforward, but now we have to be smart.

Actually, the product of all elements of A[] except A[i] is equal to the product of all elements before A[i] and those after A[i].

We can traverse A[] twice, once from left to right and once in the opposite way and multiply all the elements we find before A[i].

We’ll pretend to have a new array called Output[] to store the output of the first pass, assigning Output[i] the product of all elements preceding A[i]:

1. let rec firstpass product input =
2.     match input with
3.     | [] -> []
4.     | x::xs -> product :: firstpass (product * x) xs

For the second pass we need to move from right to left, but this can be done by reversing the input arrays and moving as usual:

1. let secondpass product input arr =
2.     let rev_input = List.rev input
3.     let rev_arr = List.rev arr
4.     let rec rev_secondpass product (input:list<int>) arr =
5.       match arr with
6.       | [] -> []
7.       | x::xs -> rev_secondpass (product * input.Head) input.Tail xs @ [(x * product)]
8.
9.     rev_secondpass product rev_input rev_arr

Both firstpass and secondpass expect an integer argument called product, which will be always be 1 at the beginning and will be used to store the intermediate products during the recursive calls.

With these functions we can just define an input array and apply them to get the result.

The following is the complete F# code:

1. let input = [ 4; 3; 2; 1; 2 ]
2.
4.
5.   let rec firstpass product input =
6.     match input with
7.     | [] -> []
8.     | x::xs -> product :: firstpass (product * x) xs
9.
10.   let secondpass product input arr =
11.     let rev_input = List.rev input
12.     let rev_arr = List.rev arr
13.     let rec rev_secondpass product (input:list<int>) arr =
14.       match arr with
15.       | [] -> []
16.       | x::xs -> rev_secondpass (product * input.Head) input.Tail xs @ [(x * product)]
17.
18.     rev_secondpass product rev_input rev_arr
19.
20.   values |> firstpass 1 |> secondpass 1 values
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