Coding Challenge

[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"

[Karl Sanford]

Again, no... the method itself is recursive, no need to call it recursively. Look at the last few lines of the method:

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

What this means is that if the mothod changed the original string in any way, it will call itself again to see if there is anything else to be removed. Otherwise, if nothing changed, then the method will return the result up the call stack.

Be The Noise

[Marc Clifton]

You're algorithm results in: "  monkey dog horse " (two leading spaces) rather than: " monkey dog horse " It's really odd how every solution that I've seen somehow fails to take this into account (as well as the fact that after the first trim, there will be a leading or trailing space in the result.) Are we that bad at reading specs, meeting simple requirements, and testing our code? :sigh: Marc

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

[Marc Clifton]

Results in too many leading spaces. The output should have one leading and one trailing space. Marc

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