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post #21 of 55

9/26/12 at 10:41am

lordikon
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Quote:

Originally Posted by RadMabbit

Quote:

Originally Posted by lordikon

Yea I'm pretty confused as well now.

I've gathered what I think are part of your requirements:
1.) Sorting an array
2.) Removing duplicates from an array
3.) Possibly using another array to keep track of something from the first array.

Yea just about, except I need to keep the order of the original array. So say I have this one
2,3,4,5,5,1,1,1
I need to remove duplicates AND keep the order from left to right, so I would get
2,3,4,5,1

So what is the second array needed for? Or do you keep the original, and make a copy of it that doesn't have dupes?

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post #22 of 55

9/26/12 at 4:56pm

tompsonn
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Quote:

Originally Posted by lordikon

Nice. I'd like to see him explain how that code works to his teacher though. Bit shifting, pointer creation/passing/arithmetic are probably things he hasn't learned yet.

Also, since he's able to use C++, not sure he should be using malloc/free/realloc, and he may not have learned that stuff yet either.

They never say in the OP what they can and can't do

When I did a prelim course in C++ a very long time a go (which I got bored of, go figure) the first thing they taught was basic C (which I already knew) and the use of C memory allocators.

Quote:

Originally Posted by RadMabbit

You got it, none of that stuff made any sense! Managed to delete the duplicates, but doing that meant I had to sort, then I lose the original order. That was the only thing I was left with....
Stupid C++

I got no problem explaining how it works

Quote:

Originally Posted by lordikon

So what is the second array needed for? Or do you keep the original, and make a copy of it that doesn't have dupes?

Second array... I am asking the same question....

--

So basically you need to sort an array and remove duplicates? It becomes easier when the array is sorted... although not much point if the order has to remain the same anyway

Edited by tompsonn - 9/26/12 at 5:43pm

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post #23 of 55

9/26/12 at 5:35pm

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RadMabbit
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The second array was a suggestion by the teacher for a way to keep the order.....so then after you have deleted the duplicates, you can put the array back into it's original order.

My teacher suggested today that there is another option by using a merge sort, by modifying it so that during the merge step, I can disregard duplicate values. This would combine the sorting/duplication together. However, this has really put me in a whole other world.

Also one thing to note, I believe the variable n, is passed into the function is the number of values in the array.

I apologize for this guys. Have not been clear at all.

post #24 of 55

9/26/12 at 5:42pm

tompsonn
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Quote:

Originally Posted by RadMabbit

The second array was a suggestion by the teacher for a way to keep the order.....so then after you have deleted the duplicates, you can put the array back into it's original order.
My teacher suggested today that there is another option by using a merge sort, by modifying it so that during the merge step, I can disregard duplicate values. This would combine the sorting/duplication together. However, this has really put me in a whole other world.
Also one thing to note, I believe the variable n, is passed into the function is the number of values in the array.
I apologize for this guys. Have not been clear at all.

Why bother sorting if you want to keep the order? It makes no sense to me....

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post #25 of 55

9/26/12 at 5:45pm

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RadMabbit
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I have no idea.....this is my dilemma. If there is a simpler way to do it, I would love to know, I just don't truly understand your method....driving me INSANE!

post #26 of 55

9/26/12 at 5:57pm

tompsonn
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Quote:

Originally Posted by RadMabbit

I have no idea.....this is my dilemma. If there is a simpler way to do it, I would love to know, I just don't truly understand your method....driving me INSANE!

The simpler way would be my way, or lordikon's way

My way works by using a bit-array. When we loop over each number in the array, I set a bit in the bit-array to 1 that corresponds to the number. For example, if there was the number 10 in the array, I would turn on (set to 1) bit 10 in the bit array.

In the loop, if the bit has already been set then it is a duplicate so I skip over it.
All the distinct values go into a new array which gets returned at the end of the function.

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post #27 of 55

9/26/12 at 6:04pm

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Nope not getting it. He was very adamant about using arrays! So lordikon's method might put me in a bad place. The teacher seems to despise C and any relation too C will receive a penalty worse then death....I was really enjoying all aspects of Computer Science until THIS guy showed up. Never had so many issues....making me question how to create an int variable.......

Im working on something now....this could be a solution, Ill post it in a moment!

post #28 of 55

9/26/12 at 6:14pm

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AHAHAHAHAHAHAHAAH YES

I am a GENIUS!!!! All praise my simplistic solution to this easy problem. tompsonn you can take you awesomely complicated code and give it to someone who is a lot smarter than me, as can you lordikon. Who needs you guys......nah im joking, but check this out...it really is a pretty cool little idea!!!

Code:

//
//  main.cpp
//  4. List Distinct
//
//  Created by Adam Thoseby on 25/09/2012.
//  Copyright (c) 2012 Adam Thoseby. All rights reserved.
//

#include <iostream>
#include <algorithm>
#include <string>
using namespace std;

int listdistinct(string b[], int n);

int main(int argc, const char * argv[])
{
    int n=4;
    string b[4] = {"4","1","4","4"};
    n = listdistinct( b, n);
}

int listdistinct(string b[], int n) {
    int i, j;
    
    /* new length of modified array */
    int nLength = 1;
    
    for(i=1; i< n; i++){
        
        for(j=0; j< nLength ; j++)
        {
            
            if(b[i] == b[j])
                break;
        }
        
        /* if none of the values in index[0..j] of array is not same as array[i],
         then copy the current value to corresponding new position in array */
        
        if (j==nLength )
            b[nLength++] = b[i];
    }
    
    for (int t=0; t<nLength; t++) {
        cout<<b[t]<<endl;
    }
    return nLength;
}

Piece of artwork right?

One thing though, is this complexity still n log(n)?

post #29 of 55

9/26/12 at 6:18pm

tompsonn
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It is technically O(n^2) because of the nested loop.

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post #30 of 55

9/26/12 at 6:23pm

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Quote:

Originally Posted by tompsonn

It is technically O(n^2) because of the nested loop.

Awesome.....why wont this 5th story window open.

Can you see what I was trying to do now? I want to be able to do something like that but have a worst case complexity of nlogn...

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