CODEWITHSUNDEEP
c++repeat
Vak Karapetian
Vak Karapetian

Reputation: 11

Need To Print Repeated Numbers Just Once / C++

Can't use libraries and other methods.

As you can see my program finds the repeated numbers and print it but I need to print the numbers just once.

As example if entered:

7 1 1 2 1 2 2 9

It should print

1 2

In case there is no any repeated number:

7 1 2 3 4 5 6 7

There should not be any output!

Also note, that the first number is the length of array:

#include <iostream>

int main() 
{
  unsigned size;
  std::cin >> size;
  int* myArray = new int[size];
  
  for (int i = 0; i < size; i++) {
      std::cin >> myArray[i];
  }

  for (int i = 0; i < size; i++) {
    bool found = false;
    for (int j = 0; j < i && !found; j++) {
        found = (myArray[i] == myArray[j]);
    }
    if (!found) {
        std::cout << myArray[i] << " ";
    } 
  }
   


   delete []myArray;
}

Upvotes: 1

Views: 757

Answers (3)

Kostiantyn Kurbatskyi
Kostiantyn Kurbatskyi

Reputation: 364

Instead of two bools as the other user suggested I would use the counter basically doing the same thing but with one variable. The trick is to check if you have already had the number you are checking right now before so that you wouldn't print it again. And then to check the rest of the list for duplicates.

#include <iostream>

int main() 
{
  unsigned size;
  std::cin >> size;
  int* myArray = new int[size];
  
  for (int i = 0; i < size; i++) {
      std::cin >> myArray[i];
  }

  for (int i = 0; i < size; i++) {
    int count = 0;
    //check if the number has already been found earlier
    for (int j = 0; j < i && !count; j++) {
        if(myArray[i] == myArray[j]) count++;
    }
    //check the rest of the array for the repeated number
    if (!count) {
       for (int j = i; j < size; j++) {
        if(myArray[i] == myArray[j]) count++;
       } 
    }
    //print if repeated
    if (count > 1) {
        std::cout << myArray[i] << " ";
    } 
  }
   
   delete []myArray;
}

Upvotes: 0

molbdnilo
molbdnilo

Reputation: 66371

The first occurrence of a repeated number has no occurrences before it and at least one after it.
This is reasonably easy to detect:

for (int i = 0; i < size; i++) {
    bool before = false;
    for (int j = 0; j < i && !before; j++) {
        before = myArray[i] == myArray[j];
    }
    if (!before) {
        bool after = false;
        for (int j = i + 1; j < size && !after; j++) {
            after = myArray[i] == myArray[j];
        }
        if (after)
        {
            cout << myArray[i] << " ";
        }
    }
}

Upvotes: 2

Mureinik
Mureinik

Reputation: 311163

The easiest approach would probably be to use a set, but I'm not sure if that's allowed under the "can't use other libraries" rule.

Using just arrays, for each item you could iterate over all the items before it, and only print it if it wasn't found there:

for (int i = 0; i < size; i++) {
    bool found = false;
    for (int j = 0; j < i && !found; j++) {
        found = (myArray[i] == myArray[j]);
    }
    if (!found) {
        cout << myArray[i] << " ";
    }
}

Upvotes: 2

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