FSharp.it

Checking whether a series of parenthesis is balanced and valid is a classic puzzle in computer science and can be easily met during a job interview.

In my opinion, there is no better solution than using a regular expression and solve the problem with a single (but almost unreadable) line of code, however I’d like to describe an algorithm which can be easily implemented in any functional or imperative programming language.

The idea is simple: start from the first character and check if it is an opening parenthesis. In this case we have to push it on a LIFO data structure such as a stack and continue with the next character.

If the character is a closing parenthesis, we take (pop) the top element from the stack and check if it makes a valid couple of brackets with the examined element.

If this is not the case, the series is not balanced, otherwise we can discard both parenthesis and go on with the next character, until the series is over.

When there are no parenthesis left to examine, if the stack is empty our series is valid; if there are still elements on the stack we can conclude that the series is unbalanced.

Lets give a look at the F# implementation I wrote to answer Dev102’s challenge:

#light

let brackets = [('(',')'); ('[',']'); ('{','}'); ('<','>')]

let is_opening bracket =  List.exists ( fun (a,b) -> a = bracket) brackets
let is_closing bracket =  List.exists ( fun (a,b) -> b = bracket) brackets
let is_valid (opening, closing) = List.exists (fun (a,b) -> (a,b) = (opening, closing)) brackets

let rec check_valid stack char_array  =
   match char_array with
   | [] -> stack = List.Empty
   | c::cs when is_opening c -> check_valid (c::stack) cs
   | c::cs ->
      match stack with
      | [] -> false
      | x::xs when is_valid (x, c) -> check_valid xs cs
      | x::xs -> false

At the beginning I defined the list of valid couples and three functions to check respectively if a char is an opening or closing bracket or if a couple is valid.

After that there is a function implementing the algorithm explained above. The stack is just a normal list, where items are always added and removed from the front.

There is an interesting series of programming job interview challenges proposed by Dev102.com, which is now at its tenth puzzle:

This week question is pretty easy. Your input is an unsorted list of n numbers ranging from 1 to n+1, all of the numbers are unique, meaning that a number can’t appear twice in that list. According to those rules, one of the numbers is missing and you are asked to provide the most efficient method to find that missing number. Notice that you can’t allocate another helper list, the amount of memory that you are allowed to allocate is O(1). Don’t forget to mention the complexity of your algorithm…

I’m not sure I understood correctly the constraint related to the memory allocation.

In my opinion, when they say we are limited to O(1), they mean that we can only allocate a single numeric variable and not any other data structure.

According to this interpretation, the solution is quite easy.

First of all, we take our only variable and store into it the sum of all the numbers between 1 and n + 1, which can be easily computed remembering that 1 + 2 + … + n = n (n + 1) / 2.

Then, we subtract each element of the array from this value, eventually the result is actually our missing number.

The functional implementation of this imperative algorithm is straightforward:

#light

let sum n = ((n + 1) * (n + 2)) / 2

let answer nums = (nums |> List.length |> sum) - (List.reduce_left (+) nums)

Let’s talk about the complexity of the algorithm.

If the length of the input list list is known, evaluating the sum of the first n natural numbers takes O(1), otherwise we have to scan the entire list, i.e. O(n).

Then we have to subtract each element of the list, and this takes another iteration, so another O(n).

Therefore we have O(n) + O(n), which is definitely O(n), and can be easily optimized a little by scanning the list only once.

The only problem left is: “am I following the instructions“?