Probelm says, "The front end is the end with the letter that comes first in the alphabet", so you have to flip car "bda" into "adb".

so in case {"acbd", "abcd", "bda"}, the answer is 1.

and if the car A and car B can be linked, A.front <= A.back == B.front <= B.back, so sort like this.

sort(cars.begin(),cars.end(),comp);

bool comp(string s1,string s2){
   if(s1.front() != s2.front()) return s1.front() < s2.front();
   if(s1.back() != s2.back()) return s1.back() < s2.back();
   return s1 < s2;
}

ardiankp's code

sorry about my English — I'm not good at it.