Match a simple balanced language

Question

I apologise if the title is not very descriptive. If someone can recommend a clearer one I'll edit it. I have an exercise question as:

The language L={aⁿbⁿ} where n ≥ 1, is a language of all words with the following properties:

• The words consist of strings of a’s followed by b’s.
• The number of b’s is equal the number of a’s.
• Examples of words that belong to L are:

ab, where n=1;

aabb, where n=2;

aaabbb, where n=3;

aaaabbbb, where n=4.

One way to test if a word w belong to this language L is to use a stack to check if the number of a’s balances the number of b’s. Use the following header and write a function isInLanguageL2 that uses a stack to test if any word belongs to L. If w belongs to L then the isInLanguageL2 should return true otherwise isInLanguageL2 should return false.

bool isInLanguageL(string w);

Note the following:

• Only words belonging to L should be accepted.
• The word bbaa does not belong to L.
• Do not count the number of a’s or b’s.

I've written the following code which solves the problem.

bool isInLanguageL2(string w){

    linkedStackType<string> wordStack;

    int size = w.length();
    int a = 0;
    int b = 0;

    if(size % 2 != 0)   //word length must be an equal number
        return false;
    else{
        //read first half of word
        for(int i = 0; i < size/2; i++){    //check if the first half consists of A's
            if(w[i] != 'a' && w[i] != 'A')
                return false;
            else
                wordStack.push(string(1, w[i])); //convert char to string and push to stack if the letter is valid
        }

        //read second half of word
        for(int i = size/2; i < size; i++){     //check if the second half consists of B's
            if(w[i] != 'b' && w[i] != 'B')
                return false;
            else
                wordStack.push(string(1, w[i]));    //convert char to string and push to stack if the letter is valid
        }
    }

    //check number of A's and B's in the stack
    while(!wordStack.isEmptyStack()){
        if(wordStack.top() == "b" || wordStack.top() == "B"){
            b++;
            wordStack.pop();
        }
        else{
            a++;
            wordStack.pop();
        }
    }

    //check if number of B's is equal to number of A's
    if(b==a)
        return true;

}

However, since the question says "Do not count the number of a’s or b’s.", I've instead created two stacks and just compared their lengths. Is there a way to do this using a single stack since the question mentions using a stack without explicitly counting the A's and B's to test the word.

bool isInLanguageL2(string w){

    linkedStackType<string> wordStackA, wordStackB;

    int size = w.length();
    int a = 0;
    int b = 0;

    if(size % 2 != 0)   //word length must be an equal number
        return false;
    else{
        //read first half of word
        for(int i = 0; i < size/2; i++){    //check if the first half consists of A's
            if(w[i] != 'a' && w[i] != 'A')
                return false;
            else
                wordStackA.push(string(1, w[i])); //convert char to string and push to stack if the letter is valid
        }

        //read second half of word
        for(int i = size/2; i < size; i++){     //check if the second half consists of B's
            if(w[i] != 'b' && w[i] != 'B')
                return false;
            else
                wordStackB.push(string(1, w[i]));    //convert char to string and push to stack if the letter is valid
        }
    }

    //check if number of B's is equal to number of A's
    if(wordStackA.length() == wordStackB.length());
        return true;

}

Also, I have a follow up to this question. A sort of extension. Should I add it here or create a new question for it as it might make this question rather long.

Answer 1

This exercise is somewhat unsettling. On the one hand, a stack, as part of a pushdown automaton, is the canonical way to decide balanced languages such as this one; on the other hand, there are much more efficient, idiomatic and practical ways to solve that particular exercise. It is unclear whether it's a computer science exercise or a language learning exercise, one to make concepts sink in or one to discover C++'s expressiveness.

A stack would have been more easily justified if the function argument had been a std::istream& for instance, since it can be traversed only once; or if starting / ending tags had come of different kinds (if they're all the same an integer makes for a perfect stack and counting shouldn't be discriminated).

As @Mike Borland said, the trick to answer that question is to avoid explicit counting by popping the stack filled with as while you iterate over the rest of the string. Something along the lines:

#include <stack>
#include <string>

bool is_ab_balanced(const std::string& str) {
    if (!str.size()) return false; 
    auto it = str.begin();
    std::stack<char> as;
    while (it != str.end() && *it == 'a') as.push(*it++);
    while (!as.empty()) {
        if (it == str.end() || *it != 'b') return false;
        as.pop(); ++it;
    }
    return it == str.end();
}

By the way, you'll notice that:

1. it always is better to use standardized containers unless you need customized behavior: std::stack is certainly better optimized and tested than linkedStackType<>

2. you shouldn't take a string by value as an argument, unless you intend to modify it while keeping the original string unchanged. Here, there's no modification, so use a const reference instead.

3. you should get rid of unused variables (see a and b in your second version). Those superfluous variables probably are a by-product of bad habits: declaring variables too far ahead of their use, and not enabling all warnings when you compile your code.

4. iterators are a better way to iterate over a range: it's more idiomatic, it's necessary to leverage the power of the STL, and it also avoids that extra size variable.

5. A few more key-strokes are better than using namespace std, which is is a bad idea

But, as I said, a good exercise should look more like real life. In real life, you use a stack only if it is the best way to go. So how would you check if a string belongs to the language if you could choose without restriction?

My take on that would be:

#include <string>
#include <algorithm>

using Iterator = std::string::iterator;
bool is_balanced(Iterator first, Iterator last) {
    auto middle = std::next(first, std::distance(first, last) / 2);
    return std::equal(first, middle, middle, last, [](auto a, auto b) {
        return a == 'a' && b == 'b';
    });
}

Or, other more meaningful exercises would have been:

• how would you check if parenthesis are balanced in an expression (I wouldn't use a stack here either)?

• how would you check if html tags are correctly balanced (here using a stack is meaningful)?

Answer 2

papagaga makes some good points.

A few things that I'd add are:

I like the initial check about string length. In terms of the intention of the underlying question, it's kind of cheating: if you're modelling a push down automaton you don't have access to the input length before you read it. Nevertheless it's the sort of quick check that is often worth having if you don't have to comply with such arbitrary restrictions: checks that can avoid doing a bunch of unnecessary work and can simplify the problem in a way that reduces the chance of hitting bugs.

You generally want to avoid using strings to hold individual characters, because it is unnecessarily slow and expensive in terms of memory. A standard c++ string holds at least the length of the string (which you don't need to hold because it's always 1 in the entries on your stack) on top of the actual string, which means you're using up at least five times as much memory as you need!

When you get a programming task the first thing you should do is check the specification for what input is actually allowed, because that informs how complicated you have to make the code. For example, I would expect from that description that the strings would not contain capital letters, and if they did then "A" would not be considered the same symbol as "a". As such, I'd avoid checking for the capital versions.

The most important thing, though, is to make sure that you understand what concept the question is trying to explore. Measuring the size of a stack is really just an expensive way of counting characters. The key thing to note about papagaga's first version is that it doesn't need a concept of how to compare integers, whether an overt b==a or the more subtle wordStackA.length() == wordStackB.length().

Answer 3

Although you already have some good remarks about your code and a short functional solution in the accepted answer, I'd like to offer an alternative procedural solution. The observation here is that you do not actually have to count the as and bs, you just have to go through the string in two directions and check that you only encounter as on one side and bs on the other. As long as the length is even (and not zero) this will give you the right answer too:

bool isInLanguageL2(string w) {
    int len = w.length();

    // Exclude the empty string and odd-length strings
    if(len == 0 || len % 2 == 1) { 
        return false; 
    }

    // Can do this with one counter until len/2, but then 
    // you need to check index i and len - 1 - i.
    for(int front = 0, back = len - 1; front < back; ++front, --back) {
        if(w[front] != 'a' || w[back] != 'b') { 
            return false; 
        }
    }

    return true;
}