What is the space complexity of below method?

Question

I would like to understand space complexity for below method. I tried simplest approach myself. Time complexity is O(n). Unable to determine whether space complexity is O(1) or O(n) as I am reassigning a (string) in every iteration.

    public boolean anagram2(String a, String b) {
        if (a.length() != b.length()) {
            return false;
        }

        for (int i = 0; i < a.length(); i++) {
            if (a.contains("" + b.charAt(i))) {
                a = a.replace("" + b.charAt(i), "");
            }
        }

        return a.length() == 0;
    }

Can you also please explain difference between space complexity and auxiliary space?

Answer 1

They are all destroyed at the end, but the space complexity would be the maximum amount of space needed at any point while the algorithm is running.

You're creating a new string for each iteration, using O(n) space. Only one of them needs to be kept in memory at a time, so the total space complexity would be O(n) .

Answer 2

As they already said, space complexity seems O(1) since no additional variables are used and, even if you reassign a string at each iteration, the previous one is not anymore tracked and gets destroyed at the end of the iteration.

Edit: After @Henry comment, I thought at it a but more and I agree with him, the space complexity seems O(n) since at each iteration you are creating a new string of variabile length in relation to the previous string and assigning it to a variable.

But, I want to point out that the time complexity is not O(n) as you said. Even if you have a single loop, the methods contains and replace aren't costing only O(1), such methods search / work on the whole string and as such, every time you use one, it costs O(a) where a is the length of the string the method is working on. For this reason, the cost is more approximable at O(n * n)

Usually for what I remember, search, replace, sum operations on strings does not cost O(1)