** UnhandledExceptionEventHandler :: OFFICIAL LC / CHATTER THREAD ** - Page 420 - Computer Technical Help

Last edited by tyler_cracker; 10-10-2013 at 12:20 AM. Reason: one regret from my 2012 job search: not managing to work into an interview the phrase "COBOL of the 2000s"

Quote

10-10-2013 , 12:32 AM

#10477

daveT

S.A.G.E. Master

Join Date: Jun 2005 Posts: 23,955

"Thanks for coming in, daveT, but.... "

Yeah, that solution probably isn't very good. I'd still prefer a candidate with a single pass across the array instead of doing double comparisons across the array.

Quote

10-10-2013 , 12:51 AM

#10478

jaytorr

centurion

Join Date: May 2003 Posts: 127

daveT, your solution is not bad at all, just in the wrong language. Are you a Lisp programmer by any chance?

I think that in java you can do the same by keeping two pointers. Or the following, which is very similar to what you posted and IMO more clear than the official solution:

Code:

public static int countRepeats(int[] a) 
  {
  if (a.length <= 1) 
    return 0;
			
  int count = 0; 
  for(int i = 1; i < a.length; i++) 
  {
    if (a[i] == a[i-1]) 
    {
      count++;
      int repeated = a[i];
      for( i++; i < a.length && a[i] == repeated; i++);
    }
  }
  return count;
}

Quote

10-10-2013 , 12:57 AM

#10479

Xhad

Carpal \'Tunnel

Join Date: Jul 2005 Posts: 10,962

Quote:

Originally Posted by jaytorr

daveT, your solution is not bad at all, just in the wrong language. Are you a Lisp programmer by any chance?

Look at his location.

(I actually almost responded with something like, "We don't all program in languages where the default data structure is a linked list")

Quote

10-10-2013 , 01:05 AM

#10480

daveT

S.A.G.E. Master

Join Date: Jun 2005 Posts: 23,955

Quote:

Originally Posted by jaytorr

daveT, your solution is not bad at all, just in the wrong language. Are you a Lisp programmer by any chance?

I think that in java you can do the same by keeping two pointers. Or the following, which is very similar to what you posted and IMO more clear than the official solution:

Code:

public static int countRepeats(int[] a) 
  {
  if (a.length <= 1) 
    return 0;
			
  int count = 0; 
  for(int i = 1; i < a.length; i++) 
  {
    if (a[i] == a[i-1]) 
    {
      count++;
      int repeated = a[i];
      for( i++; i < a.length && a[i] == repeated; i++);
    }
  }
  return count;
}

Yeah, that's pretty close to what I was thinking of, except that repeated would probably be initialized outside of the loop so that the if is more like if(repeated == a[i]]), but I'm not sure how well that would work in Java land.

I'm not sure what the second for loop is doing. Care to enlighten me?

Quote:

Originally Posted by Xhad

Look at his location.

(I actually almost responded with something like, "We don't all program in languages where the default data structure is a linked list")

Then it wouldn't surprise you to know that I thought of recursion first.

Quote

10-10-2013 , 01:27 AM

#10481

jaytorr

centurion

Join Date: May 2003 Posts: 127

It's just incrementing the pointer while a[i] == repeated. It's the equivalent of

Code:

array = array[1:] until candidate != array[0]

Quote

10-10-2013 , 01:49 AM

#10482

daveT

S.A.G.E. Master

Join Date: Jun 2005 Posts: 23,955

Ah, I see, except that the array isn't getting getting shortened, just pointing to the next index. I was thinking more along the lines of popping the top of the stack / dequeue, but I wasn't sure how to represent it. I implied copying when I didn't really mean to. Thanks for that food for thought.

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10-10-2013 , 02:51 AM

#10483

gaming_mouse

Carpal \'Tunnel

Join Date: Oct 2004 Posts: 13,786

Quote:

Originally Posted by DirtySprite

Anyway my teacher had a solution which I think is a bit more elegant

no, it's not.

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10-10-2013 , 04:22 AM

#10484

clowntable

Carpal \'Tunnel

Join Date: Jun 2006 Posts: 45,557

Quote:

Originally Posted by tyler_cracker

well array slices are for people who...

yeah i can't do it. lol java, lol java.util.Arrays.copyOfRange, lol camelCase.

My hate for Java has gotten less and less over the years. To the point where I'm actually thinking about properly relearning it. NeoCobol is fairly accurate though lol but then again there's Android

I think what's helping it is even if you don't like Java the language the JVM is pretty nice.

Quote

10-10-2013 , 05:01 AM

#10485

gaming_mouse

Carpal \'Tunnel

Join Date: Oct 2004 Posts: 13,786

Quote:

Originally Posted by Xhad

I don't think there's any convenient way to get array slices in Java

EDIT: oh, I just found it right after I posted:

http://stackoverflow.com/questions/1...-array-in-java

Doubt it's faster or even reasonably close b/c of all the copying overhead

I think one of the other answers is actually the more "java" way to do it:

Code:

List<MyClass> subArray = Arrays.asList(array).subList(index, array.length);

This will not copy the underlying array. Instead, we are essentially wrapping it with a decorator, adding to it more "list"-like behavior. The subList call is now a method on your object, so you are actually programming with an OO style rather than applying a global utility function to your array.

Spoiler:

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10-10-2013 , 08:19 AM

#10486

gaming_mouse

Carpal \'Tunnel

Join Date: Oct 2004 Posts: 13,786

Also, I just realized I somehow missed the obvious (and I think best) solution of just counting the number of changes to get the number of unique values, then subtracting that from the total:

Code:

    public static int countRepititions(int[] a) {
        
        if (a.length <= 1) return 0;

        int numChanges = 0;
        for(int i=1; i<a.length; i++) {
            if (a[i] != a[i-1])
                numChanges++;
        }   
        int numUniques = numChanges +1;
        return a.length - numUniques;       
    }

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10-10-2013 , 09:19 AM

#10487

kerowo

lolcat

Join Date: Nov 2005 Posts: 37,190

1, 2, 2, 2, 3, 3 has 2 changes but you would return 3 (if my paper computer is right). Also, isn't it counting elements with a count > 2? You are only counting changes not the number of elements that match.

Quote

10-10-2013 , 09:22 AM

#10488

Gullanian

Carpal \'Tunnel

Join Date: Dec 2006 Posts: 14,014

This one should work some of the time:

Code:

public static int countRepititions(int[] a) {
    return new Random().nextInt(a.length/2)
}

Quote

10-10-2013 , 09:27 AM

#10489

kerowo

lolcat

Join Date: Nov 2005 Posts: 37,190

Would it be faster to look for array[i] != array[i+1] && array[i+1]==array[i+2]?

1 2 2 3 3 4 4 5 6
1t2t2 = 1
2f2f3 = 0
2t3t3 = 2
3f3f4 = 0
3t4t4 = 3
4f4f5 = 0
4t5f6 = 0

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10-10-2013 , 10:12 AM

#10490

gaming_mouse

Carpal \'Tunnel

Join Date: Oct 2004 Posts: 13,786

Quote:

Originally Posted by kerowo

right, 3 is the answer (as i understand it).
there are 2 extra twos and 1 extra 3, for a total of 3 repititions

Quote

10-10-2013 , 10:14 AM

#10491

Gullanian

Carpal \'Tunnel

Join Date: Dec 2006 Posts: 14,014

I interpreted the question as being 2, there are 2 distinct numbers that occur more than once

Quote:

Example given is for the input { 1; 1; 1; 2; 4; 4; 5; 6; 6 } the answer would be 3 because three different numbers (1,4,6) appear more than once.

Quote

10-10-2013 , 10:27 AM

#10492

gaming_mouse

Carpal \'Tunnel

Join Date: Oct 2004 Posts: 13,786

Quote:

Originally Posted by Gullanian

I interpreted the question as being 2, there are 2 distinct numbers that occur more than once

yeah you are right. i somehow misremembered the question between yesterday and today. but your interpretation is how my previous answer and the teacher's answer work, so that's correct. so... nm my latest post

Quote

10-10-2013 , 10:44 AM

#10493

tyler_cracker

Carpal \'Tunnel

Join Date: Apr 2005 Posts: 15,828

Quote:

Originally Posted by clowntable

I think what's helping it is even if you don't like Java the language the JVM is pretty nice.

i agree with this. in the early 2000s when perl6 was still vaporware[1] and people were crowing about ParrotVM[2], i was like who gives a ****? why do i care that my interpreted language gets compiled to some bytecode situation before going off to the hardware.

nowadays, the rise of meaningfully useful languages which leverage the jvm (e.g. clojure) have proven me wrong. turns out larry wall knows more than me about language design[3].

[1] O WATE

[2] cwidt?

[3] drudgesiren.gif

Quote

10-10-2013 , 11:31 AM

#10494

kerowo

lolcat

Join Date: Nov 2005 Posts: 37,190

I guess the code would look like this for my above idea:

Code:

public static int checkRepetition(int[] a)
	{
		int repetitionCount = 0;
	
		for(int i=0; i<(a.length-2); i++)
		{
			if (a[i] != a[i+1] && a[i+1]==a[i+2]
	                {
                              repetitionCount++;
                         }
		}		
		return repetitionCount;	
	}

Quote

10-10-2013 , 12:07 PM

#10495

gaming_mouse

Carpal \'Tunnel

Join Date: Oct 2004 Posts: 13,786

That is basically like my solution from yesterday: http://forumserver.twoplustwo.com/sh...ostcount=10468

except you're looking forward and you miss the edge case. For example that code fails on:

int[] a = {1,1,2,2,3,4,4,6};

Quote

10-10-2013 , 03:23 PM

#10496

jaytorr

centurion

Join Date: May 2003 Posts: 127

While we're still skinning this cat, here are a couple more ways.

First a modified version of what I posted earlier. It's forward looking and that eliminated the need for the initial array length check - the main loop does the check. Also changed the for loop that increments the pointer to a while loop to make it more clear.

Code:

public static int countRepeats(int[] a) 
{
  int repeated, count = 0; 
  for(int i = 0; i < a.length-1; ) 
  {
	if (a[i] == a[++i]) 
	{
	  count++;
	  repeated = a[i++];
	  while(i < a.length-1 && a[i] == repeated) i++;
	}
  }
  return count;
}

Can someone with more Java/C experience comment on the use of postfix and prefix operators above? I'm learning C++ at the moment and see them all the time in others' code, but I'm not sure how idiomatic they are. I'm particularly leery about the line:

Code:

if (a[i] == a[++i])

Here's a partially recursive version (it still has a while loop).

Code:

public static int countRepeats3_r(int[] a, int start, int end, int count)
{
  if (end-start <= 1) 
    return count;
  if (a[start] == a[start+1]) 
  {
    int next = start+2;
    while(next < end && a[next] == a[start]) next++;
    return countRepeats3_r(a, next, end, count+1);
  }
  else
    return countRepeats3_r(a, start+1, end, count);
}	
	
public static int countRepeats3(int[] a) 
{
    return countRepeats3_r(a, 0, a.length, 0);
}

I tried a fully recursive version, but it was a bit ugly for my liking.

Quote

10-10-2013 , 03:31 PM

#10497

daveT

S.A.G.E. Master

Join Date: Jun 2005 Posts: 23,955

If you really want to show off, do:

A = evenElements(array)
B = oddElements(array)
ans = len(intersection(A, B))

Quote

10-10-2013 , 04:38 PM

#10498

daveT

S.A.G.E. Master

Join Date: Jun 2005 Posts: 23,955

In case its not clear, A and B are sets.

Quote

10-10-2013 , 04:52 PM

#10499

derada4

old hand

Join Date: Feb 2011 Posts: 1,544

What's the consensus on C#?

I'm a CS major graduating this winter and just got a job offer doing C# on a .NET stack. I really like the company and the people I'd be working with, the salary is pretty much just what I was looking for, but working in a non open source language is slightly less than ideal to me. Although I'm a noob so I figured I'd ask here.

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10-10-2013 , 05:17 PM

#10500

clowntable

Carpal \'Tunnel

Join Date: Jun 2006 Posts: 45,557

C# the language is fairly well designed but Windows sucks (imo) and the entire environment while theoretically independent of OS (there's Mono for Linux etc.) is not particularly awesome on other operating systems. If you end up working there and are in the environment anyways make sure to check out F# as a side project...neat language.

C# and .net are technically* open but surrounded by closed stuff.
*iirc the MS implementation doesn't exactly follow the open standard but it's been a while since I used C#.

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