Help me resolve this issue . This is a long challenge problem

general
, ,
1 
                             #include <iostream>
                   #include <cmath>
                 // 3
         //  xy    1 1 
      //  xyxxy 3 2
        //  yy    0 2

          int main()
         {
               int t;
            std::cin >> t;
         while(t--)
        {
          std::string s;
        std::cin >> s;
           int count1=0 ;
           int count2=0;
          for (int i=0;i<s.length();++i)
             {
                if(s[i]=='x')
                   ++count1;

              else
          ++count2;
        }

            if(count1==0 || count2==0)
            {
            std::cout << 0 <<'\n';
           }
       else
              {
             std::cout << floor((count2+count1)/2) << '\n';
  
            }

     }
       return 0;


     }
2

Your logic seems to be completely wrong. You cannot depend on the count of x and y. You need to take care of their positions too.
If and only if x is adjacent to y or vice-versa, you increase the pair count by 1.
AC code:

#include <bits/stdc++.h>

using namespace std;

int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    while (t--)
    {
        string s;
        cin >> s;
        long long p = 0;
        for (int i = 0; i < s.size()-1; ++i)
        {
            if ((s[i] == 'x' && s[i+1] == 'y') || (s[i] == 'y' && s[i+1] == 'x'))
            {
                p += 1;
                ++i;
            }
        }
        cout << p << '\n';
    }
}
3

Thanks !