Myujth

I want to count all the different string elements in an array.
So my input would be:

5 Lemon Orange Lemon Mango Lemon

And the output should be this:

3

The problem with my code is, that my code counts all the elements, and not just the different and I can't figure out why.

Here is my code:

#include <iostream>

using namespace std;

int main()

 int N; 
 cin >> N;
 string Tname;
 string data[N];
 int counter = 0;

 for(int i = 0; i<N; i++)
 
 cin >> Tname;
 data[i] = Tname;
 

 for(int l = 0; l<N; l++)
 
 int k = 0;
 while(k<N && (data[l] != data[k]))
 
 k++;
 
 if(k<N)
 
 counter += 1;
 

 
 cout << counter << endl;
 return 0;

The problem is algorithmic: every item is equal to itself, which will end your k loop prematurely. In addition, you only increment when the item is repeated.

I propose you to change the loops, so not to compare every items with every other items, but only items, to items previously processed:

for(int l = 0; l<N; l++)

 int k = 0;
 while(k<l && data[l] != data[k]) // only previous items
 
 k++;
 
 if(k==l) // if no identical, we can add this one
 
 cout<<l<<" "<<data[l]<<endl;
 counter += 1;
 

Not related: variable length arrays are not legal C++ even if some mainstream compilers accept it. I'd suggest to use a vector to emulate this feature: vector<string> data(N);

Online demo

if i understand well your problem , you want the value that has the max appearances in the array, some modifications are needed to achieve this :

#include <iostream>

using namespace std;

int main()

 int N; 
 cin >> N;
 string Tname;
 string data[N];
 int counter = 0;

 for(int i = 0; i<N; i++)
 
 cin >> Tname;
 data[i] = Tname;
 

 int tempCounter; // a temporary counter for each item of the array .
 for(int l = 0; l < N; l++)
 
 tempCounter = 0;
 int k = 0;
 while(k<N)
 
 if(data[l] == data[k])
 tempCounter++;
 k++;
 
 if(tempCounter > counter) // if the new counter is higher than the counter 
 counter = tempCounter;
 
 cout << counter << endl;
 return 0;

the last if should be if(k+1==N)

because you all the time stop the while before the k reach N

and k must start from l

Your logic is, you add 1 to the counter if it is not in the list's remaining part
But the code check the full list so you never count thw world whitch in the list twice.