#include <iostream>
#include <cstdlib>
#include <deque>

using namespace std;

int main()
{
    int h;
    char s[200001];
    scanf("%s", &s);
    int n=strlen(s);
    deque<char> v (s, s + n  );

    if(v.size()==2  && v[0]==v[1] ) return 0;
    b: h=2;
    for(int i=0; i<v.size()-1 ;i++)
          if(v[i] == v[i+1])
        {
            v.erase(v.begin() + i);
            v.erase(v.begin() + i);
            h=1;
        }

    if(h==1) goto b;

    for(int i=0; i<v.size(); i++)
        cout << v[i];

        return 0;
}

Hello,

If you are still working on this, let you know that it can't be solved with erase.
I implemented a similar algorithm in Java which goes through the string and deletes the successful characters and it also got TLE on test 6.The easiest way is removing the successful characters as you read them.
nc->new character from the input.

if(i!=-1&&c[i]==nc){//new char is equal to the char at the end of the string, remove it
         i--;
     }else{ //add the new char to the string
         i++;
         c[i]=nc;
     }
Hope this helps

You got TLE because deque has poor deletion performance on positions other than the front and back.

This helped me, thanks.