Coding Challenge

[Simon_Whale]

No as you are removing all instances of dog and cat from the string and not from either end of the string

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[Slacker007]

Perl, perhaps. One line, maybe two.

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[Chris Maunder]

You going to code or are you going to nitpick? ;)

cheers, Chris Maunder The Code Project | Co-founder Microsoft C++ MVP

[Bassam Abdul Baki]

No to be pedantic about things, bit if you're tokenizing based on "dog" and "cat", your final answer should be "  monkey dog horse " (begins with 2 spaces) or "monkey dog horse " (begins with no space). Otherwise, the requirement on what to do with spaces is incomplete. :)

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[JacquesDP]

static void Main(string[] args)
{
string testString = "dog cat monkey dog horse dog";
List splitString = new List(testString.Trim().Split(new char[]{' '}));
for (int index = 0; index < splitString.Count; index++)
{
if (splitString[0].Trim().Equals("dog") || splitString[0].Trim().Equals("cat"))
splitString.RemoveAt(0);
else if (splitString[splitString.Count - 1].Trim().Equals("dog") || splitString[splitString.Count - 1].Trim().Equals("cat"))
splitString.RemoveAt(splitString.Count - 1);
else
break;
index = 0;
}
string final = "";
foreach (var entry in splitString)
final += entry + " ";
Console.WriteLine(final.Trim());
Console.ReadKey();
}

No matter how long he who laughs last laughs, he who laughs first has a head start!

[Chris Maunder]

The challenge is to remove the strings that are provided. Nothing is said about removing (or, indeed, caring about, whitespace)

cheers, Chris Maunder The Code Project | Co-founder Microsoft C++ MVP

[Lost User]

Well, I was working from the assumption (or maybe wishful thinking!) that Chris is actually intending to use this to parse tags and remove language keywords[^]. Best Wishes, -David Delaune

[NormDroid]

Failed: 1. No formatting 2. Language Dependent 3. Too verbose

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[NormDroid]

Shit my eyes are bleeding :)

Software Kinetics Wear a hard hat it's under construction
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[Lost User]

Hmmm, Do you think it would be uglier and more complicated than this[^] dual lambda? :-D Best Wishes, -David Delaune

[Chris Maunder]

No consideration of whitespace is provided.

cheers, Chris Maunder The Code Project | Co-founder Microsoft C++ MVP

[Chris Maunder]

No whitespace treatment is specified in the spec so you can't assume tokenisation is possible

cheers, Chris Maunder The Code Project | Co-founder Microsoft C++ MVP

[Lost User]

Awww shucks. :sigh:

[Chris Maunder]

Won't work if you have "dog dog text". It will only remove the first "dog"

cheers, Chris Maunder The Code Project | Co-founder Microsoft C++ MVP

[Mark_Wallace]

Too easy.

While DogsAndCatsAreStillInTheLine
LineEnd = "left"
GetRidOfCatsAndDogs(LineEnd)
LineEnd = "right"
GetRidOfCatsAndDogs(LineEnd)
Wend

GetRidOfCatsAndDogs {
OpenCanOfDogFood(ThisEnd)
OpenCanOfCatFood(ThisEnd)
}

It's so simple that it's not even worth optimising it or using recursion. For future updates, though, it would be a good idea to use a set for the animals to remove, so that different ones could be included, another for the line ends (so that multiple lines could be serviced), and a third for the food types -- although getting bamboo shoots for pandas, eucalyptus leaves for koalas, and starlets for great whites might cause difficulties in implementation.

I wanna be a eunuchs developer! Pass me a bread knife!

[Chris Maunder]

Thank you, Fermat.

cheers, Chris Maunder The Code Project | Co-founder Microsoft C++ MVP

[Lost User]

My approach is to use substring to grab the current end and starting strings. Substring is also used to update the result. I am sure there are some faster ways, but this approach is very clean IMO.

public string Parse(string input, List remove)
{
string result = input.Trim();
bool startRemoved;
bool endRemoved;

do
{
    startRemoved = false;
    endRemoved = false;

    foreach (var match in remove)
    {            
        //Determine if the remove string is fitable with in the remaining output
        bool inRange = result.Length >= match.Length;

        //Capture the beginning and ending characters
        string beginning = result.Substring(0, match.Length);
        string end = result.Substring(result.Length - match.Length, match.Length);

        if (inRange && beginning.Equals(match))
        {
            //Update the result with the matched section removed
            result = result.Substring(match.Length, result.Length - match.Length).TrimStart();
            startRemoved = true;
        }

        if (inRange && end.Equals(match))
        {
            //Update the result with the matched section removed
            result = result.Substring(0, result.Length - match.Length).TrimEnd();
            endRemoved = true;
        }
    }
} while (startRemoved || endRemoved);
return result;        

}

[Edit] Added a do while. The do while shuts down if nothing is removed. If anything is removed it must go back and check the removal collection. Optimization could be to track which is removed to then check up to that point. E.g. "dog" "cat" and "monkey" are to be removed (and collection is in that order). If "cat" is removed (start or end), then "monkey" will be checked on active foreach. "dog" and "cat" are the only ones that need be checked on the next do while iteration (unless one of them removes another character set) [Edit] Added Trimmers.

Computers have been intelligent for a long time now. It just so happens that the program writers are about as effective as a room full of monkeys trying to crank out a copy of Hamlet.

[Chris Maunder]

Who said anything about tokenising?

cheers, Chris Maunder The Code Project | Co-founder Microsoft C++ MVP

[Chris Maunder]

For Dalek Dave's benefit, I dub this the Most Unique Solution.

cheers, Chris Maunder The Code Project | Co-founder Microsoft C++ MVP

[Nagy Vilmos]

No substrings [except for the end] or tokenizing in Java:

public static String trimStrings(String string, String\[\] trimmings) {
    if (trimmings == null || trimmings.length == 0) {
        return string.trim();
    }
    int left = 0;
    int right = string.length()-1;

    boolean trimmed;
    do {
        while (left < right && string.charAt(left) == ' ') {
            left++;
        }
        while (left < right && string.charAt(right) == ' ') {
            right--;
        }

        trimmed = false;
        for (int trim = 0; trim < trimmings.length; trim++) {
            if (string.indexOf(trimmings\[trim\], left) == left) {
                left += trimmings\[trim\].length();
                trimmed = true;
            } else if (string.lastIndexOf(trimmings\[trim\], right) ==
                    (right - trimmings\[trim\].length() + 1)) {
                right -= trimmings\[trim\].length();
                trimmed = true;
            }
        }
    } while (trimmed);

    return string.substring(left, right+1);
}

---

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