Coding Challenge

[Simon_Whale]

Private Sub Button2_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button2.Click
Dim Input As String = "dog dog monkey dog horse cat cat dog"
Dim ToRemove As String() = New String() {"dog", "cat"}

    For Each Value As String In ToRemove
        While True
            If Input.StartsWith(Value) = True Then
                Input = Input.Substring(Value.Length + 1)
            ElseIf Input.EndsWith(Value) = True Then
                Input = Input.Substring(0, Input.LastIndexOf(Value) - 1)
            Else
                Exit While
            End If
        End While
    Next

    MessageBox.Show(Input)
End Sub

Lobster Thermidor aux crevettes with a Mornay sauce, served in a Provençale manner with shallots and aubergines, garnished with truffle pate, brandy and a fried egg on top and Spam - Monty Python Spam Sketch

[hairy_hats]

Collin Jasnoch wrote:

perceiver

Which is defined as what? The cat is inside the box, observing its own state, so no external observer is needed to collapse the waveform.

[hairy_hats]

:-D An alive cat knows it's alive so the waveform has already collapsed to "alive".

[Chris Losinger]

it ain't elegant, but it appears to work:

#include <string>
#include <vector>
#include <algorithm>

using namespace std;

bool len (const string& s1, const string& s2)
{
return s1.length() > s2.length();
}

string trim_right(const string &source , const string& t = " ")
{
string str = source;
return str.erase( str.find_last_not_of(t) + 1);
}

bool findLeft(const char*p, size_t start, vector<string> &srch, size_t & foundLen)
{
foundLen = 0;
for (vector<string>::iterator it=srch.begin();it!=srch.end();it++)
{
if (strncmp(p + start, (*it).c_str(), (*it).length())==0)
{
foundLen = (*it).length();
return true;
}
}
return false;
}

size_t findRight(const char*p, size_t start, vector<string> &srch, size_t & foundLen)
{
foundLen = 0;
for (vector<string>::iterator it=srch.begin();it!=srch.end();it++)
{
if (strncmp(p + start - (*it).length(), (*it).c_str(), (*it).length())==0)
{
foundLen = (*it).length();
return true;
}
}
return false;
}

string stripper(string inp, vector<string> srch)
{
string ret="";

const char *p = inp.c_str();

size_t left = 0;
size_t right = inp.length();
size_t foundLen = 0;

// descending len sort. for greedy matching
std::sort(srch.begin(), srch.end(), len);

// from left
bool matched=false;
do
{
matched=false;
if (findLeft(p, left, srch, foundLen))
{
matched = true;
left+=foundLen;
while (isspace(*(p + left)) && *(p+left)) left++;
}
} while (*(p+left) && matched);

if (left==inp.length()) return ret;

// from right
do
{
matched=false;
if (findRight(p, right, srch, foundLen))
{
matched = true;
right-=foundLen;
while (isspace(*(p+right)) && right>=left) right--;
}
} while (right >= left && matched);

ret = inp.substr(left, right - left);

ret = trim_right(ret); //?

return ret;
}

int _tmain(int argc, _TCHAR* argv[])
{
vector<string> srch;
srch.push_back("doggie");
srch.push_back("dog");
srch.push_back("cat");

stripper("doggiedog cathorse cat", srch);

}

[Andy Brummer]

If spaces aren't counted as special characters the output should be " cat monkey dog horse " because cat isn't at the end after removing dog.

Curvature of the Mind now with 3D

[Michael Bergman]

Can I used delegates? If so, I will delegate this to one of my co-workers. That should do it!

m.bergman

For Bruce Schneier, quanta only have one state : afraid.

To succeed in the world it is not enough to be stupid, you must also be well-mannered. -- Voltaire

Honesty is the best policy, but insanity is a better defense. -- Steve Landesberg

[Lost User]

This will not remove all of the strings. You will only remove the first instances of them if the string starts with it or ends with it. If your collection is {"dog", "cat"} But your string is "cat dog monkey" you will remove cat but not dog.

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.

[Lost User]

It is defined as the one holding the experiment. The cat is the experiment and therefore can not be the observer. Any experiement needs observation externally or it is not sound.

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.

[NormDroid]

Internationalization (Uk, German etc.)

Software Kinetics Wear a hard hat it's under construction
Metro RSS

[NormDroid]

Can be done in RegEx albeit terse, I've just been experimenting I do believe somebody with day-to-day contact with reg ex will achieve this.

Software Kinetics Wear a hard hat it's under construction
Metro RSS

[Karl Sanford]

Umm, no, it would remove both cat and dog. Just ran your example using my code, and it returned " monkey"... maybe you should try the same... [hint] The solution is recursive [/hint]

Be The Noise

[Lost User]

http://www.codeproject.com/Lounge.aspx?msg=4118816#xx4118816xx[^]

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.

[Sentenryu]

why no one used Regex? ok, this code is far from optimum, but i had no time to do it better :(

using System;
using System.Text.RegularExpressions;

namespace chalenge
{
class Program
{
public static void Main(string[] args)
{
string input = "dog cat dog monkey dog horse dog";
string[] removes = { "dog", "cat" };
bool hasMore = false;
do{
hasMore = false;
for(int i =0; i < removes.Length; i++){

				while (Regex.IsMatch(input, @"^\\s\*" + removes\[i\] + @"\\s\*.\*")){
					input = Regex.Replace(input, @"(?^\\s\*)" + removes\[i\] + @"(?\\s\*.\*)", "${w1}${w2}");
				}
				
				while (Regex.IsMatch(input, @".\*\\s\*" + removes\[i\] + @"\\s\*$")){
					input = Regex.Replace(input, @"(?.\*\\s\*)" + removes\[i\] + @"(?\\s\*$)", "${w1}${w2}");
				}
				
				foreach(var item in removes){
					if(Regex.IsMatch(input, @"^\\s\*" + item + @"\\s\*.\*") || Regex.IsMatch(input, @".\*\\s\*" + removes\[i\] + @"\\s\*$")){
						hasMore = true;
						break;
					}
				}
				
			}
		}while(hasMore);
		Console.Write(input);
		Console.ReadKey();
	}
}

}

(sorry by the english, i'm brasilian...)

[Lost User]

Yeah cos there are no other words with different meanings. You don't have to like it but they are both valid definitions. There were trees down either side of the road.

Every man can tell how many goats or sheep he possesses, but not how many friends.

[hairy_hats]

Quantum collapse occurs due to quantum interactions with something else ("observation"). There is no need for that interaction to be conscious or human, it just means that one quantum system is disturbed through interaction with something else - such as a cat. :)

[Chris Maunder]

The specs say nothing of trimming whitespace, and the hardcoding of "dog" and "cat" means the solution can't be reused.

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

[Lost User]

Yes if you call it recursively you will remove it. That was not clear in your OP though...

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.

[Marc Clifton]

Nagy Vilmos wrote:

I was about to invite you into my esteemed LinkIn network, but I'm not sure after reading that.

Specs are like the uncertainty principle. The more you spec something, the more inaccurate you will be in some other area. ;) Marc

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

[PIEBALDconsult]

ChrisElston wrote:

There were trees down either side of the road.

That makes no sense -- which side? The other side I hope.

[Xiangyang Liu]

As philosophers would say: "That depends on how you define the concept of END". :-D

My Younger Son & His "PET"