Check my anagram code from a job interview in the past.

Posted by Michael Dorgan on Stack Overflow See other posts from Stack Overflow or by Michael Dorgan
Published on 2010-04-25T03:31:20Z Indexed on 2010/04/25 3:33 UTC
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Had the following as an interview question a while ago and choked so bad on basic syntax that I failed to advance (once the adrenalin kicks in, coding goes out the window.)

Given a list of string, return a list of sets of strings that are anagrams of the input set. i.e. "dog","god", "foo" should return {"dog","god"}. Afterward, I created the code on my own as a sanity check and it's been around now for a bit. I'd welcome input on it to see if I missed anything or if I could have done it much more efficiently. Take it as a chance to improve myself and learn other techniques:


void Anagram::doWork(list input, list> &output)
{
    typedef list> SortType;
    SortType sortedInput;

// sort each string and pair it with the original
for(list<string>::iterator i = input.begin(); i != input.end(); ++i)
{
    string tempString(*i);
    std::sort(tempString.begin(), tempString.end());
    sortedInput.push_back(make_pair(*i, tempString));
}

// Now step through the new sorted list
for(SortType::iterator i = sortedInput.begin(); i != sortedInput.end();)
{
    set<string> newSet;

    // Assume (hope) we have a match and pre-add the first.
    newSet.insert(i->first);

    // Set the secondary iterator one past the outside to prevent
    // matching the original
    SortType::iterator j = i;
    ++j;

    while(j != sortedInput.end())
    {
        if(i->second == j->second)
        {
            // If the string matches, add it to the set and remove it
            // so that future searches need not worry about it
            newSet.insert(j->first);
            j = sortedInput.erase(j);
        }
        else
        {
            // else, next element
            ++j;
        }
    }

    // If size is bigger than our original push, we have a match - save it to the output
    if(newSet.size() > 1)
    {
        output.push_back(newSet);
    }

    // erase this element and update the iterator
    i = sortedInput.erase(i);
}

}

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