Coding Challenge

[Bassam Abdul Baki]

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

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[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 ; .

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[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:

[jesarg]

I'll be the first to put up the recursive solution in C#.

private string TrimCatDog(string input)
{
if (input.StartsWith(" ") || input.EndsWith(" "))
return TrimCatDog(input.Trim());
if (input.StartsWith("cat") || input.StartsWith("dog"))
return TrimCatDog(input.Substring(3));
if (input.EndsWith("cat") || input.EndsWith("dog"))
return TrimCatDog(input.Substring(0, input.Length - 3));
return input;
}

[Manfred Rudolf Bihy]

Marc Clifton wrote:

There's probably a regex solution, but that's gross.

Glad to be of service to you ;P

using System;
using System.Collections.Generic;
using System.Text.RegularExpressions;

namespace StripWords
{
static class Program
{
static void Main()
{
String input, replace;
input = replace = "dog cat monkey dog horse dog";
String stripThese = "dog cat";
String[] words = stripThese.Split(new Char[] { ' ' }, StringSplitOptions.RemoveEmptyEntries);
String strRegEx = String.Empty;
foreach (String word in words)
{
strRegEx = strRegEx + String.Format(@"^\s*{0}\s*|\s*{0}\s*$|", word);
}
strRegEx = strRegEx.Trim(new char[] { '|' });
Regex trimRegExp = new Regex(strRegEx);

        bool replaceNeeded = false;
        do
        {
            if (replaceNeeded = trimRegExp.Match(replace).Length > 0)
            {
                replace = trimRegExp.Replace(replace, "");
            }
        } while (replaceNeeded);
        Console.WriteLine("Trimming \\"{0}\\" from input \\"{1}\\" yields \\"{2}\\"!", stripThese, input, replace);
        Console.ReadLine();
    }
}

}

Cheers! :)

"With sufficient thrust, pigs fly just fine."

Ross Callon, The Twelve Networking Truths, RFC1925

[Marc Clifton]

Collin Jasnoch wrote:

but this approach is very clean IMO.

But won't work, right? The desired result is " monkey dog horse " Did you notice the leading and trailing spaces? ;) Plus, your result calculation doesn't take into account that the result will have a leading or trailing space in it, just resulting in a failure for the next calculation of beginning and end. Marc

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

[Marc Clifton]

FAIL! ;) You're not preserving the leading and/or trailing space in the result after removing a token from the beginning or end. But cool none-the-less, and probably solvable, right? Marc

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

[Lost User]

Every sofware project starts with making a plan with strict schedules about deliverables and defining the architectural artefacts. We need to set up a scrum meeting about choosing the appropriate platform and team composition. We could do this sometimes next week, please send me your suggestions matching your schedules. Attached is the invoice for the 25% of the estimated project costs.

[Bassam Abdul Baki]

Does the Stripper animate the removal of text or does it drop them all at once? :-D BTW, I assume those are CP's tab sizes and not spaces. One hamster needs to be shot for such wide tabs.

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[PIEBALDconsult]

Now remove "the cat" from the string.

[Karl Sanford]

Here's my stab at a solution in C#: [Edit: Chris updated the guidance, and I changed the method accordingly.]

public static string CPtrimmer(string input, string[] trims, StringComparison compare = StringComparison.InvariantCultureIgnoreCase)
{
String Input = input;

foreach (var trimWord in trims)
{
    if (Input.StartsWith(trimWord, compare))
        Input = Input.Remove(Input.IndexOf(trimWord, compare), trimWord.Length);
    if (Input.EndsWith(trimWord, compare))
        Input = Input.Remove(Input.LastIndexOf(trimWord, compare), trimWord.Length);
}

if (Input != input)
    return CPtrimmer(Input, trims);
else
    return Input;

}

[modified] Given the exchange below, I am including the usage of this method as well (console C#):

static void Main(string[] args)
{
String input = "dog cat monkey dog horse dog";
String[] trims = new String[] { "dog", "cat" };

String result = CPtrimmer(input, trims);
// After CPtrimmer has returned, the result is " cat monkey dog horse "

}

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