counting duplicates and put the unique entries in to new array in array [c++]

Question

q.Write a program that contains a function named "remove_duplicates" that takes an array of integers in random order, and then eliminates all the duplicate integers in the array. The function should take three arguments:

An array of integers which is (read and filled in main)
The size of the array.
A variable passed by reference to be printed in main to show how many values in the array without duplication. The function should not return a value, but if any duplicate integers are eliminated, then the function should count that, so the new value tells the number of distinct integers in the array. Suppose the array passed to the function is as shown below, and the size of the array passed to the function is 10.

0 1 2 3 4 5 6 7 8 9 58 | 26 | 91 | 26 | 70 | 70 | 91 | 58 | 58 | 66

The function should change the array to look like this:

0       1       2     3       4      5      6      7     8      9    
58 | 26 | 91 | 70 | 66 | ?? | ?? | ?? | ?? | ??

and it should change the value of the distinct counter so that it is 5. The question marks in the cells after the 5th cell indicate that it does not matter what numbers are in those cells when the function returns.

something is wrong with my function

void remove_duplicates ( int h[] ,int n ,int &count ){
count = 0;
int a[100];
for (int i = 0 ; i < n  ; i++ )
        for(int j = i ; j<n ; j++){
            if(h[i]!=h[j])
                        a[i]=h[i];
            else
            count++;

do you get an error message from the compiler? It is missing some '}' — gbulmer
– gbulmer, Commented Apr 13, 2012 at 11:30
Is order important? std::set probably won't maintain order. — Open AI - Opting Out
– Open AI - Opting Out, Commented Apr 13, 2012 at 14:02

Anon Mail · Accepted Answer · 2012-04-13 12:30:59Z

1

Copy the entire array into a std::set. The set will eliminate the duplicates, meaning insertion of duplicates will fail.
Copy the set back to the array.
The number of unique values is the size of the set.

answered Apr 13, 2012 at 12:30

Anon Mail

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1 Comment

N_A Over a year ago

Using a set is probably the fastest way to do this.

AlexTheo · Accepted Answer · 2012-04-13 12:27:07Z

0

There are many solutions of these problem I will show you one without temp arrays. If your requirements allow you to use the sorting algorithm then sort the array before applying this code:

unsigned int numberOfDuplicates = 0;
for( unsigned int i = 1; i<numberOfElements; i++ )
{
  if( array[i-1] == array[i] ){
    numberOfDuplicates++;
    for( unsigned int j = i; j< numberOfElements-1; j++ )
    {
       array[j] = array[j+1];
    }
  }
}

Each time you find a new duplication you eat it by shifting the rest of the array one position left. A second solution is by allocating a new array and passing there only the values which are not exists already in it after that just copy the allocated array back.

Without sorting the algorithm could be:

unsigned int numberOfDuplicates = 0;
for( unsigned int i = 1; i<numberOfElements; i++ )
{
 for( unsigned int k = i; k < numberOfElements; k++){
   if( array[i-1] == array[k] ){
      numberOfDuplicates++;
      for( unsigned int j = k; j< numberOfElements-1; j++ )
      {
       array[j] = array[j+1];
      }
   }
  }
}

edited Apr 13, 2012 at 12:27

answered Apr 13, 2012 at 11:52

AlexTheo

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1 Comment

juanchopanza Over a year ago

I think sorting the array isn't compatible with OP's requirements.

Adina T. · Accepted Answer · 2012-04-13 12:31:24Z

Your count will be larger than it should when the function is done, because you increment it more than once for each element. You should try something like this:

    void remove_duplicates ( int h[] ,int n ,int &count ){
count = n;
int a[100],ok,p;
for (int i = 0 ; i < count-1  ; i++ ) {
    ok=1;
    for(int j = i+1 ; j<count ; j++)
        if(h[i]==h[j]) {
            ok=0;
            p=j;
        }
    if(!ok) {
        for(int j = p+1; j<count; j++) 
            h[j-1]=h[j];
        count--;
    }
}
count=n-count;
}

"ok" checks if v[i] can be found somewhere else in the array and p is the position where it appears, so it knows which element to eliminate. I assumed you need to keep the first appearance of an element in the array, otherwise you could have eliminated it easier. Hope it helps.

Spidey · Accepted Answer · 2012-04-13 12:45:55Z

0

If you have a defined range or a possible invalid value (say, -1), you can set the duplicates to this invalid value, and upon finding a duplicate while you traverse the array, you shift elements back to the start.

answered Apr 13, 2012 at 12:45

Spidey

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