Finding 3 or more duplicates in array - c++

Question

I need write a program which checks if there are 3 or more matching numbers in an array. My code works fine until there is an array like "2 2 3 3 5 5 4 4 1 1", then it approves that there are 3 or more duplicates in the array which is not true. Maybe someone knows a simple solution which would help me? Or do I need to overwrite my code?
Here is my code:
#include <iostream>
using namespace std;
void funk(int n, int a[], int &kiek);
int main()
{
int n, a[101],kiek=0;
cin >> n;
for (int i = 0; i < n; i++) {
cin >> a[i];
}
funk(n, a, kiek);
if (kiek > 2) {
cout << "TAIP";
}
else
cout << "NE";
}
void funk(int n, int a[], int &kiek)//funkcijos kūnas
{
int j;
for (int i = 0; i < n; i++) {
for (j = i + 1; j < n; j++) {
if (a[i] == a[j])
kiek++;
cout << kiek;
}
}
}
This is the input:
10
2 2 3 3 5 5 4 4 1 1
This is the output I need to get:
NE

Answer 1

The problem that your code has is:
You are comparing any 2 numbers and never reset the counter. So if there is an 1 1, you are increasing the counter. If there is then an 2 2 you are also increasing the counter. And for the final 3 3 you also increase the same counter. Then it is 3. Although there are only 2 same values. This can never work. What you could do is
Read all values
Count the frequency of EACH different value
Check if the frequency of the values
If any of the counts is greater than 2, then show corresponding message
Show count of each value
I will show you the "more-modern" C++ approach and will use C++ algorithms for the below example solution.
First we will get the number of values to work with from the user. We will store the values in a std::vector. And, we use std::copy_n to copy the values from std::cin to our std::vector. For that we will use the std::istream_iterator that will iterate over the elements given by the user. So, we use a simple one-liner to read all values from the user.
Next is the frequency counting. For this we have a standard solution in C++. You will find it in dozens of places in the net. We will use a std::map. The key is the integer that we read into the vector and the value is the counter. With the std::map's index operator [] we add a value to the map, if it is not yet existing. With the ++ we simply do the counting, whether the value was already in the std::map or has just been added. Also this is a very simple one-liner.
Then, we check, if any of the counts is greater than 2. For this we will use the STL algorithm std::any:of with a very simple lambda. With that, we can create your desired result.
Last, but not least, we show all values and their count, if the count is greater than 2. This we do with an ultra simple range based for loop. We extract the values out of the counter-std::map using structered bindings.
Please see:
#include <iostream>
#include <map>
#include <vector>
#include <algorithm>
#include <iterator>
int main() {
// Give instructions
std::cout << "How many values do you want do read? Please specify: ";
// Read the number of values to enter
size_t numberOfValues{ 0U }; std::cin >> numberOfValues;
// Read the given number of values from std::cin
std::vector <int> values(numberOfValues);
std::copy_n(std::istream_iterator<int>(std::cin), numberOfValues, values.begin());
// Count each value
std::map<int, size_t> counter{};
std::for_each(values.begin(), values.end(), [&counter](const int& i) { counter[i]++; });
// Check, if any count is bigger than 2
if (std::any_of(counter.begin(), counter.end(), [](const std::pair<int, size_t> & c) { return c.second > 2; }))
std::cout << "TAIP\n";
else
std::cout << "NE\n";
// Look, if there is any number with a count bigger than 2 and show the result
for (const auto& [value, count] : counter)
if (count > 2) std::cout << value << " --> " << count << "\n";
return 0;
}
I hope this gives you an idea on how this could be done . . .