Coding Challenge

[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);
}

---

Panic, Chaos, Destruction. My work here is done. Drink. Get drunk. Fall over - P O'H OK, I will win to day or my name isn't Ethel Crudacre! - DD Ethel Crudacre I cannot live by bread alone. Bacon and ketchup are needed as well. - Trollslayer Have a bit more patience with newbies. Of course some of them act dumb - they're often *students*, for heaven's sake - Terry Pratchett

[realJSOP]

Being a programmer, I cannot resist a coding challenge. Do not misconstrue this to be a general lifting of my personal CodeProject boycott. Usage:

string text = "dog cat dog doghorse monkey dog fishcat dog cat";
string[] removers = new string[] {"cat", "dog"};
do
{
} while (Challenge(ref text, removers));

Method:

//-----------------------------------------------------------------------
private bool Challenge(ref string text, string[] removeThese)
{
bool changed = false;
for (int i = 0; i < removeThese.Length; i++)
{
string target = removeThese[i];
int start = -1;
int end = -1;
do
{
end = (text.StartsWith(target)) ? target.Length : -1;
if (end > -1)
{
text = text.Substring(end);
changed = true;
}
text = text.Trim();
} while (end > -1);
do
{
start = (text.EndsWith(target)) ? text.Length - target.Length : -1;
if (start > -1)
{
text = text.Substring(0, start);
changed = true;
}
text = text.Trim();
} while (start > -1);
}
return changed;
}

You can also easily make this a recursive method.

".45 ACP - because shooting twice is just silly" - JSOP, 2010
-----
You can never have too much ammo - unless you're swimming, or on fire. - JSOP, 2010
-----
"Why don't you tie a kerosene-soaked rag around your ankles so the ants won't climb up and eat your candy ass." - Dale Earnhardt, 1997

[Bassam Abdul Baki]

Still, shouldn't you end up with two spaces at the start if you're just removing "dog" and "cat"?

Web - BM - RSS - Math - LinkedIn

[Lost User]

Norm .net wrote:

2. Language Dependent

Nothing stated you could not use a specific language. In fact part of the point it seemed was it can be done so many different ways with different languages...

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.

[hairy_hats]

In that case, shouldn't "cat dog monkey..." end up as "  monkey..." with two leading spaces? (It might just be the HTML rendering changing two to one in your OP.)

[vonb]

At least my version works. I know it can be done in much fewer lines, but the concept is there.

[hairy_hats]

It might be the site's HTML rendering which reduces two or more adjacent spaces to one. (I put about 10 in the last sentence but it doesn't display that way!)

[Jason Hooper]

Obligatory comic: http://www.sonnyradio.com/whatthecustomerreallywanted.htm[^]

Jason

[Bassam Abdul Baki]

Which is why I used & nbsp ; .

Web - BM - RSS - Math - LinkedIn

[Dalek Dave]

I think it shows he has a terrific sense of humour!

------------------------------------ I will never again mention that I was the poster of the One Millionth Lounge Post, nor that it was complete drivel. Dalek Dave CCC Link[^] Trolls[^]

[Marc Clifton]

Dang, what are we, a bunch of whiners? It's a PROGRAMMING CHALLENGE! START YOUR CODING ENGINES!!! Do we need detail specs? Hell NO!!! There's probably a regex solution, but that's gross. Here's something I whipped together in about 15 minutes (took a bit of debugging, a few nuances to it) that I hope readable and somewhat reusable:

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace Stripper
{
public enum StringPoint
{
None,
Start,
End,
}
public static class StringHelpersExtensions
{
public static string RightOf(this String src, string s)
{
string ret = src;
int idx = src.IndexOf(s);
if (idx != -1)
{
ret = src.Substring(s.Length);
}
return ret;
}
public static string LeftOfRightmostOf(this String src, string s)
{
string ret = src;
int idx = src.LastIndexOf(s);
if (idx != -1)
{
ret = src.Substring(0, idx);
}
return ret;
}
public static bool StartsOrEndsWith(this String src, string[] items, out string match, out StringPoint whichEnd)
{
bool ret = false;
whichEnd = StringPoint.None;
match = String.Empty;
foreach (string item in items)
{
if (src.StartsWith(item))
{
match = item;
whichEnd = StringPoint.Start;
ret = true;
break;
}
if (src.EndsWith(item))
{
match = item;
whichEnd = StringPoint.End;
ret = true;
break;
}
}
return ret;
}
}
class Program
{
static void Main(string[] args)
{
string input = "dog cat monkey dog horse dog";
string[] stripOf = { "dog", "cat" };
string desiredOutput = " monkey dog horse ";
string result = Stripper(input, stripOf);
if (result == desiredOutput)
{
Console.WriteLine("Success");
}
else
{
Console.WriteLine("Fail! '" + result + "'");
}
}
static string Stripper(string input, string[] stripOf)
{
string ret = input;
string match;
StringPoint whichEnd;
string test = ret.Trim();
string leftPad = String.Empty;
string rightPad = String.Empty;
while (test.StartsOrEndsWith(stripOf, out match, out whichEnd))
{
switch (whichEnd)
{
case StringPoint.Start:
// The result always preserves the leading space separating the token, so add it back in.
leftPad = " ";
ret = leftPad + test.RightOf(match).Trim() + rightPad;
break;
case StringPoint.End:
// The result always preserves

[PIEBALDconsult]

ChrisElston wrote:

'each of two.

That is "both", not "either". Don't use a word for two meanings when there is already a word for the other. You want to ride a see-saw -- you may sit at either end.

[Marc Clifton]

Did you actually test this? The resulting string needs to preserve any leading and/or trailing spaces after removing the token. Marc

My Blog
An Agile walk on the wild side with Relationship Oriented Programming

[PIEBALDconsult]

A dead cat knows it's dead? :confused: