Question

Suppose you have a dictionary that contains valid words.

Given an input string with all spaces removed, determine whether the string is composed of valid words or not.

You can assume the dictionary is a hashtable that provides O(1) lookup.

Some examples:

helloworld-> hello world (valid)
isitniceinhere-> is it nice in here (valid)
zxyy-> invalid

If a string has multiple possible parsings, just return true is sufficient.

The string can be very long. Hence think an algorithm that is both space & time efficient.

Answer 1

I think the set of all strings that occur as the concatenation of valid words (words taken from a finite dictionary) form a regular language over the alphabet of characters. You can then build a finite automaton that accepts exactly the strings you want; computation time is O(n).

For instance, let the dictionary consist of the words {bat, bag}. Then we construct the following automaton: states are denoted by 0, 1, 2. Edges: (0,1,b), (1,2,a), (2,0,t), (2,0,g); where the triple (x,y,z) means an edge leading from x to y on input z. The only accepting state is 0. In each step, on reading the next input sign, you have to calculate the set of states that are reachable on that input. Given that the number of states in the automaton is constant, this is of complexity O(n). As for space complexity, I think you can do with O(number of words) with the hint for construction above.

For an other example, with the words {bag, bat, bun, but} the automaton would look like this:

Supposing that the automaton has already been built (the time to do this has something to do with the length and number of words 🙂 we now argue that the time to decide whether a string is accepted by the automaton is O(n) where n is the length of the input string.
More formally, our algorithm is as follows:

1. Let S be a set of states, initially containing the starting state.
2. Read the next input character, let us denote it by a.
3. For each element s in S, determine the state that we move into from s on reading a; that is, the state r such that with the notation above (s,r,a) is an edge. Let us denote the set of these states by R. That is, R = {r | s in S, (s,r,a) is an edge}.
4. (If R is empty, the string is not accepted and the algorithm halts.)
5. If there are no more input symbols, check whether any of the accepting states is in R. (In our case, there is only one accepting state, the starting state.) If so, the string is accepted, if not, the string is not accepted.
6. Otherwise, take S := R and go to 2.

Now, there are as many executions of this cycle as there are input symbols. The only thing we have to examine is that steps 3 and 5 take constant time. Given that the size of S and R is not greater than the number of states in the automaton, which is constant and that we can store edges in a way such that lookup time is constant, this follows. (Note that we of course lose multiple ‘parsings’, but that was not a requirement either.)
I think this is actually called the membership problem for regular languages, but I couldn’t find a proper online reference.