Programming Forums > Application Development > C++

mychorhizzae · Mar 26th, 2008, 12:27 AM

Hi, I've been lurking here for a while now and decided to make an account to get help. The program I'm trying to write is supposed to:
Read in 20 numbers,
each of which is between 10 and 100, inclusive. As each number is read in, validate
it and store it in the array only if it is not a duplicate of a number already read.
After reading all the values, display only the unique values that the user has
entered.

I have the program functionally working except I don't know what to do for the duplicate cases. Is there an easy way to skip the value entered if duplicate?

(Toggle Plain Text)

#include <iostream>
using namespace std;
#include <iomanip>




int main() {
    int idx = 0;
    int list[20];
    int n;
    int i;
    
    cout << "Enter 20 integers between 10 and 100: " << endl;
    
    for (i=0; i<20; i++){ // user inputs 20 numbers
       
    cin >> n;
    
    if (n < 10 || n > 100){  // input must be between 10 and 100
    cout << "invalid entry" << endl;
}
            list[idx] = n;  //sets each number to the next element in array
              ++idx;

}
    for ( int j = 0; j < 10; j++ )
      cout << list[ j ] << " ";  // print numbers in array
        
    system("pause");
    return 0;
}

lectricpharaoh · Mar 26th, 2008, 1:14 AM

The easiest (but not the most efficient) method is to iterate through the array, checking to see if the number is present. If so, don't add it. You can write a function for this:

 C++ Syntax (Toggle Plain Text) 
bool isValueInArray(int value, int *array, int numValues)
{
  for(int x=0; x<numValues; ++x)
    if(array[x] == value)
      return true;
  return false;
}
bool isValueInArray(int value, int *array, int numValues)
{
  for(int x=0; x<numValues; ++x)
    if(array[x] == value)
      return true;
  return false;
}

Now, when you get the number from the user, just call this function first in the body of your 'get input from user' loop, and only add it if it returns false:

 C++ Syntax (Toggle Plain Text) 
if(isValueInArray(n, list, idx)
{
  list[idx] = n;
  ++idx;
}
if(isValueInArray(n, list, idx)
{
  list[idx] = n;
  ++idx;
}

Semantically identical to the above, by using postfix increment:

 C++ Syntax (Toggle Plain Text) 
if(isValueInArray(n, list, idx)
   list[idx++] = n;
if(isValueInArray(n, list, idx)
   list[idx++] = n;

Of course, you don't need to write a separate function for this; I did it only for clarity. You can merge it into the loop with a flag variable and a nested loop.

You're also not printing out all the valid values the user entered, unless this happens to be 10. Try making your printing loop like so:

 C++ Syntax (Toggle Plain Text) 
for(int j=0; j<idx; ++j)
  cout << list[j] << ' ';
for(int j=0; j<idx; ++j)
  cout << list[j] << ' ';

On another note, your code would be much clearer with consistent indentation, and (in my opinion, though I'm sure others will disagree) using the 'opening brace at line start' style, rather than the old K&R style. I had to scan your code a bit to tell where blocks began and ended, whereas if it were done the way I suggest, it'd have been much easier.

Sane · Mar 26th, 2008, 6:12 AM

There's a neat trick you can use to do this.

Use an array of zeroes. For each number they enter, set the n'th element to 1. After you're done, print out all elements that are 1.

This automatically stops itself from printing out duplicates, because setting an element to 1 twice still keeps it at 1.

Also, your validation of each number did not stop it from being added to the array. The assignment specifies to include duplicates as one of the 20 numbers, but do not count invalid numbers as one of the 20.

(Toggle Plain Text)

int main() {
    int idx = 0;
    int list[101] = {0}; // make an array of 101 zero's
    int n;
    int i;
    
    cout << "Enter 20 integers between 10 and 100: " << endl;
    
    for (i=0; i<20; i++) { // user inputs 20 numbers
       
        do {
            cout << "Enter number " << i+1 << ": ";
            cin >> n;
        } while (n < 10 || n > 100);  // input must be between 10 and 100

        list[n] = 1;  // this tells us the n'th number has been used

    }
    
    for ( int j = 10; j <= 100; j++ )
        if(list[j] == 1)
            cout << j << " ";  // index of all numbers that have a 1
        
    system("pause");
    return 0;
}

Note that this does not mantain the order the numbers were inputted.

If you need to mantain the order, you could keep your original code, and then use this method to output the array. As you're outputting it, set the element to 1. Then make sure you don't output a number if the list is set to 1 for that value.

Ooble · Mar 26th, 2008, 6:46 AM

This sounds like a homework assignment to me, so I'm not sure if you'd be allowed to do this, but it would be really easy with sets. Sets are a type of data structure which are always ordered and can only store unique values, which sounds like what you're looking for. There's a set class in the STL.

To initialise a set:

(Toggle Plain Text)

set<int> list;

To put stuff in:

(Toggle Plain Text)

list.insert(n);

Don't worry about putting stuff in twice - it'll just ignore the duplicates.
And to get stuff out:

(Toggle Plain Text)

for (set<int>::iterator i = list.begin(); i != list.end(); i++) {
    cout << *i << " ";
}

lectricpharaoh · Mar 26th, 2008, 7:39 PM

I made a boo-boo:

Quote:

Originally Posted by lectricpharaoh

 C++ Syntax (Toggle Plain Text) 
if(isValueInArray(n, list, idx)
{
  list[idx] = n;
  ++idx;
}
if(isValueInArray(n, list, idx)
{
  list[idx] = n;
  ++idx;
}

You need to invert that condition, and I missed a parenthesis:

(Toggle Plain Text)

if(!isValueInArray(n, list, idx))
{
  list[idx] = n;
  ++idx;
}

lectricpharaoh · Mar 26th, 2008, 7:40 PM

Double post, oops.

Mar 26th, 2008, 12:27 AM	#1
mychorhizzae Newbie Join Date: Mar 2008 Posts: 1 Rep Power: 0	duplicate numbers in arrays Hi, I've been lurking here for a while now and decided to make an account to get help. The program I'm trying to write is supposed to: Read in 20 numbers, each of which is between 10 and 100, inclusive. As each number is read in, validate it and store it in the array only if it is not a duplicate of a number already read. After reading all the values, display only the unique values that the user has entered. I have the program functionally working except I don't know what to do for the duplicate cases. Is there an easy way to skip the value entered if duplicate? (Toggle Plain Text) #include <iostream> using namespace std; #include <iomanip> int main() { int idx = 0; int list[20]; int n; int i; cout << "Enter 20 integers between 10 and 100: " << endl; for (i=0; i<20; i++){ // user inputs 20 numbers cin >> n; if (n < 10 \|\| n > 100){ // input must be between 10 and 100 cout << "invalid entry" << endl; } list[idx] = n; //sets each number to the next element in array ++idx; } for ( int j = 0; j < 10; j++ ) cout << list[ j ] << " "; // print numbers in array system("pause"); return 0; } #include <iostream> using namespace std; #include <iomanip> int main() { int idx = 0; int list[20]; int n; int i; cout << "Enter 20 integers between 10 and 100: " << endl; for (i=0; i<20; i++){ // user inputs 20 numbers cin >> n; if (n < 10 \|\| n > 100){ // input must be between 10 and 100 cout << "invalid entry" << endl; } list[idx] = n; //sets each number to the next element in array ++idx; } for ( int j = 0; j < 10; j++ ) cout << list[ j ] << " "; // print numbers in array system("pause"); return 0; }

Mar 26th, 2008, 1:14 AM	#2
lectricpharaoh Caffeinated Neural Net Join Date: Jun 2005 Location: Dry west coast of Canada Posts: 927 Rep Power: 4	Re: duplicate numbers in arrays The easiest (but not the most efficient) method is to iterate through the array, checking to see if the number is present. If so, don't add it. You can write a function for this: C++ Syntax (Toggle Plain Text) bool isValueInArray(int value, int array, int numValues) { for(int x=0; x<numValues; ++x) if(array[x] == value) return true; return false; } bool isValueInArray(int value, int array, int numValues) { for(int x=0; x<numValues; ++x) if(array[x] == value) return true; return false; } Now, when you get the number from the user, just call this function first in the body of your 'get input from user' loop, and only add it if it returns false: C++ Syntax (Toggle Plain Text) if(isValueInArray(n, list, idx) { list[idx] = n; ++idx; } if(isValueInArray(n, list, idx) { list[idx] = n; ++idx; } Semantically identical to the above, by using postfix increment: C++ Syntax (Toggle Plain Text) if(isValueInArray(n, list, idx) list[idx++] = n; if(isValueInArray(n, list, idx) list[idx++] = n; Of course, you don't need to write a separate function for this; I did it only for clarity. You can merge it into the loop with a flag variable and a nested loop. You're also not printing out all the valid values the user entered, unless this happens to be 10. Try making your printing loop like so: C++ Syntax (Toggle Plain Text) for(int j=0; j<idx; ++j) cout << list[j] << ' '; for(int j=0; j<idx; ++j) cout << list[j] << ' '; On another note, your code would be much clearer with consistent indentation, and (in my opinion, though I'm sure others will disagree) using the 'opening brace at line start' style, rather than the old K&R style. I had to scan your code a bit to tell where blocks began and ended, whereas if it were done the way I suggest, it'd have been much easier. __________________ A man's knowledge is like an expanding sphere, the surface corresponding to the boundary between the known and the unknown. As the sphere grows, so does its surface; the more a man learns, the more he realizes how much he does not know. Hence, the most ignorant man thinks he knows it all. - L. Sprague de Camp

Mar 26th, 2008, 6:12 AM	#3
Sane Programming Guru Join Date: Apr 2005 Posts: 1,799 Rep Power: 5	Re: duplicate numbers in arrays There's a neat trick you can use to do this. Use an array of zeroes. For each number they enter, set the n'th element to 1. After you're done, print out all elements that are 1. This automatically stops itself from printing out duplicates, because setting an element to 1 twice still keeps it at 1. Also, your validation of each number did not stop it from being added to the array. The assignment specifies to include duplicates as one of the 20 numbers, but do not count invalid numbers as one of the 20. (Toggle Plain Text) int main() { int idx = 0; int list[101] = {0}; // make an array of 101 zero's int n; int i; cout << "Enter 20 integers between 10 and 100: " << endl; for (i=0; i<20; i++) { // user inputs 20 numbers do { cout << "Enter number " << i+1 << ": "; cin >> n; } while (n < 10 \|\| n > 100); // input must be between 10 and 100 list[n] = 1; // this tells us the n'th number has been used } for ( int j = 10; j <= 100; j++ ) if(list[j] == 1) cout << j << " "; // index of all numbers that have a 1 system("pause"); return 0; } int main() { int idx = 0; int list[101] = {0}; // make an array of 101 zero's int n; int i; cout << "Enter 20 integers between 10 and 100: " << endl; for (i=0; i<20; i++) { // user inputs 20 numbers do { cout << "Enter number " << i+1 << ": "; cin >> n; } while (n < 10 \|\| n > 100); // input must be between 10 and 100 list[n] = 1; // this tells us the n'th number has been used } for ( int j = 10; j <= 100; j++ ) if(list[j] == 1) cout << j << " "; // index of all numbers that have a 1 system("pause"); return 0; } Note that this does not mantain the order the numbers were inputted. If you need to mantain the order, you could keep your original code, and then use this method to output the array. As you're outputting it, set the element to 1. Then make sure you don't output a number if the list is set to 1 for that value. __________________ Waterloo's Canadian Computing Competition (CCC) - Stage 2 Problems, Solutions, and Test Data

Mar 26th, 2008, 6:46 AM	#4
Ooble I eat cake for breakfast. Join Date: Jul 2004 Location: In my box. Posts: 4,434 Rep Power: 9	Re: duplicate numbers in arrays This sounds like a homework assignment to me, so I'm not sure if you'd be allowed to do this, but it would be really easy with sets. Sets are a type of data structure which are always ordered and can only store unique values, which sounds like what you're looking for. There's a set class in the STL. To initialise a set: (Toggle Plain Text) set<int> list; set<int> list; To put stuff in: (Toggle Plain Text) list.insert(n); list.insert(n); Don't worry about putting stuff in twice - it'll just ignore the duplicates. And to get stuff out: (Toggle Plain Text) for (set<int>::iterator i = list.begin(); i != list.end(); i++) { cout << i << " "; } for (set<int>::iterator i = list.begin(); i != list.end(); i++) { cout << i << " "; } __________________ Me :: You :: Them

Mar 26th, 2008, 7:40 PM	#6
lectricpharaoh Caffeinated Neural Net Join Date: Jun 2005 Location: Dry west coast of Canada Posts: 927 Rep Power: 4	Re: duplicate numbers in arrays Double post, oops. __________________ A man's knowledge is like an expanding sphere, the surface corresponding to the boundary between the known and the unknown. As the sphere grows, so does its surface; the more a man learns, the more he realizes how much he does not know. Hence, the most ignorant man thinks he knows it all. - L. Sprague de Camp

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Programming Forums > Application Development > C++

duplicate numbers in arrays