Question

I was asked about this question. I can only think of a O(nm) algorithm if n is the length of the 1st string and m is the length of the 2nd string.

Answer 1

Well, you can do it in O(n + m). Just create a reference table showing whether character exists in first string. Something like this (pseudo-code in no particular language)

// fill the table
for (int i = 0; i < a.length; ++i) {
    characterExists[a[i]] = true;
}

// iterate over second string
for (int i = 0; i < b.length; ++i) {
    if !characterExists[b[i]] {
        // remove char (or do whatever else you want)
    }
}